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SOLUTION 


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University  of  California. 


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With  the  respectful  compliments  of  the  author. 

ADDRESS, 

MR.  R.  BALLARD, 

MALVERN, 


ENGLAND. 


THE  SOLUTION 


PYRAMID  PROBLEM 


OR, 


PYRAMID   DISCOVERIES. 


WITH   A 

NEW  THEORY  AS  TO  THEIR  ANCIENT  USE. 


BY 

ROBERT  BALLARD, 

M.  INST.  C.E.,  ENGLAND  ;    M.  AMER.  SOC.  C.E. 

CHIEF    ENGINEER    OF   THE   CENTRAL   AND    NORTHERN    RAILWAY   DIVISION 
OF  THE    COLONY  OF  QUEENSLAND. 
AUSTRALIA.  • 


OP  TUB 
UNIVERSITY 


NEW   YORK: 

JOHN    WILEY  &    SONS» 

1882. 


COPYRIGHT, 

1882, 

BY  JOHN  WILEY  &  SONS. 
2/076 


PRESS  OF  j.  j.  LITTLE  &  co.f 

10    TO    20    ASTOR    PLACE,    NEW    YORK. 


NOTE. 


IN  preparing  this  worlf  for  publication  I  have  received  valuable 
help  from  the  following  friends  in  Queensland  :— 

E.  A.   DELISSER,    L.S.  and   C.E.,  Bogantungan,  who  assisted 
me  in    my  calculations,  and  furnished    many  useful    sugges- 
tions. 

J.  BRUNTON  STEPHENS,  Brisbane,  who  persuaded  me  to  pub- 
lish my  theory,  and  who  also  undertook  the  work  of  correc- 
tion for  the  press. 

J.  A.  CLARKE,  Artist,  Brisbane,  who  contributed  to  the  Illus- 
trations. 

LYNE  BROWN,  Emerald, — (photographs). 

F.  ROTHERY,  Emerald, — (models). 

and— A.  W.  VOYSEY,  Emerald, —  (maps  and  diagrams). 


CONTENTS. 


§i.     The  Ground  Plan  of  the  Gi'zeh  Group 9 

Plan  Ratios  connected  into  Natural  Numbers 13 

§2.     The  Original  Cubit  Measure  of  the  Gi'zeh  Group 15 

§5.     The  Exact  Measure  of  the  Bases  of  the  Pyramids 19 

§4.     The  Slopes,  Ratios   and  Angles  of  the   Three  Principal 

Pyramids  of  the  Gi'zeh  Group 20 

§5.     The  Exact  Dimensions  of  the  Pyramids 23 

§6.     Geometrical  Peculiarities  of  the  Pyramids 28 

§6A.  The  Casing  Stones  of  the  Pyramids 32 

§7.     Peculiarities  of  the  Triangles  3,  4,  5  and  20,  21,  29 35 

§8.     General  Observations 38 

§9.     The  Pyramids  of  Egypt  the  Theodolites  of  the  Egyptian 

Land  Surveyors 41 

§  i p.     How  the  Pyramids  were  made  use  of 44 

§n.     Desoription  of  the  ancient  Portable  Survey  Instrument. .  51 

Table  to  explain  Figure  60 64 

§12.     Primary  Triangles  and  their  Satellites; — or  the  Ancient 
System  of  Right-angled  Trigonometry  unfolded  by  a 

Study  of  the  Plan  of  the  Pyramids  of  Gizeh 65 

Table  of  some  "  Primary  Triangles"'  and  their  Satellites.  74 
§13.     The  Size  and   Shape  of  the   Pyramids  indicated  by  the 

Plan 77 

§14.     A  Simple  Instrument  for  laying  off  "Primary  Triangles"  79 

§  1 4 A.   General  Observations 80 

§15.     Primary  Triangulation 82 

§16.     The  Pentangle  or  Five-pointed  Star  the  Geometric  Sym- 
bol of  the  Great  Pyramid 91 

Table  shewing  the  comparative  Measures  of  Lines  ...  96 
§17.     The  manner  in  which  the  Slope  Ratios  of  the  Pyramids 

were  arrived  at '. 1 04 


LIST   OF   WORKS   CONSULTED. 


Penny  Cyclopedia.     (Knight,  London.  '  1833.) 

Sharpe  s  Egypt. 

u  Our    Inheritance    in    the    Great    Pyramid."      Piazzi 

Smyth. 
"  The  Pyramids  of  Egypt"     R.  A.   Proctor*-    (Article 

in  Gentleman  s  Magazine.     Feb.  1880.) 
"  Traite  de  la  Grandeur  et  de  la  Figure  de  la  Terre" 

C as  sin  i.     (A  msterdam.      1723.) 
"  Pyramid  Facts  and  Fancies."     J.  Bonwick. 


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THE 

SOLUTION  OF  THE  PYRAMID  PROBLEM. 


WITH  the  firm  conviction  that  the  Pyramids  of  Egypt 
were  built  and  employed,  among  other  purposes,  for 
one  special,  main,  and  important  purpose  of  the  great- 
est utility  and  convenience,  I  find  it  necessary  before  I 
can  establish  the  theory  I  advance,  to  endeavor  to  de- 
termine the  proportions  and  measures  of  one  of  the 
principal  groups.  I  take  that  of  Gizeh  as  being  the 
one  affording  most  data,  and  as  being  probably  one  of 
the  most  important  groups. 

I  shall  first  try  to  set  forth  the  results  of  my  inves- 
tigations into  the  peculiarities  of  construction  of  the 
Gizeh  Group,  and  afterwards  show  how  the  Pyramids 
were  applied  to  the  national  work  for  which  I  believe 
they  were  designed. 

§    i.     THE  GROUND  PLAN  OF  THE  GIZEH  GROUP. 

I  find  that  the  Pyramid  Cheops  is  situated  on  the 
acute  angle  of  a  right-angled  triangle — sometimes  called 
the  Pythagorean,  or  Egyptian  triangle — of  which  base, 
perpendicular,  and  hypotenuse  are  to  each  other  as  3, 
4,  and  5.  The  Pyramid  called  Mycerinus,  is  situate  on 

9 


10 


SOLUTION   OF   THE    PYRAMID   PROBLEM. 


the  greater  angle  of  this  triangle,  and  the  base  of  the 
triangle,  measuring  three,  is  a  line  due  east  from  Myce- 
rinus,  and  joining  perpendicular  at  a  point  due  south  of 
Cheops.  (See  Figure  i.) 


Jytycerinua 

I  find  that  the  Pyramid  Cheops  is  also  situate  at  the 
acute  angle  of  a  right-angled  triangle  more  beautiful 
than  the  so-called  triangle  of  Pythagoras,  because  more 
practically  useful.  I  have  named  it  the  20,  21,  29  trian- 
gle. Base,  perpendicular,  and  hypotenuse  are  to  each 
other  as  twenty,  twenty-one,  and  twenty-nine. 

The  Pyramid  Cephren  is  situate  on  the  greater  angle 
of  this  triangle,  and  base  and  perpendicular  are  as  be- 
fore described  in  the  Pythagorean  triangle  upon  which 
Mycerinus  is  built.  (See  Fig.  2.) 


Cheops  ' 


Ceph 


re/i 


Cephren 


JVtyeertnus. 


Cheops. 


Figure  3     represents    the   combination, — A     being 
Cheops,  F  Cephren,  and  D  Mycerinus. 

Lines  DC,  CA,  and  AD  are  to  each  other  as  3,  4, 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  II 

and  5  ;  and  lines  FB,  BA,  and  AF  are  to  each  other  as 
20,  21,  and  29. 

The  line  CB  is  to  BA,  as  8  to  7  ;  The  line  FH  is 
to  DH,  as  96  to  55  ;  and  the  line  FB  is  to  BC,  as  5  to  6. 

The  Ratios  of  the  first  triangle  multiplied  by  forty- 
five,  of  the  second  multiplied  by  four,  and  the  other 
three  sets  by  twelve,  one,  and  sixteen  respectively,  pro- 
duce the  following  connected  lengths  in  natural  num- 
bers for  all  the  lines. 

.DC 135 

CA 1 80 

AD.         .225 


FB.... ..    80 

BA ...    84 

AF 1 16 

CB 96 

'BA 84 

FH 96 

55. 


FB 8q 

BC 96 

Figure  4  connects  another  pyramid  of  the  group — 
it  is  the  one  to  the  southward  and  eastward  of  Cheops. 


12 


SOLUTION   OF  THE   PYRAMID   PROBLEM. 


In  this  connection,  A  Y  Z  A  is  a  3,  4,  5  triangle,  and 
B  Y  Z  O  B  is  a  square, 

Lines  YA  to  C A  are  as  i  to  5 
CY  to  YZ          as  3  to  i 
FO  to  ZO          as  8  to  3 
and   DA  to  AZ         as  15  to  4. 

I  may  also  point  out  on  the  same  plan  that  calling" 
the  line  FA  radius,  and  the  lines  BA  and  FB  sine  and 
co-sine,  then  is  YA  equal  in  length  to  versed  sine  of 
angle  AFB. 

This  connects  the  20,  21,  29  triangle  FAB  with  the 
3,  4,  5  triangle  AZY. 

I  have  not  sufficient  data  at  my  disposal  to  enable 
me  to  connect  the  remaining  eleven  small  pyramids  to 
my  satisfaction,  and  I  consider  the  four  are  sufficient  for 
my  purpose. 


Of  these  natural  -\ 
numbers  thebases[ . 
cf  the  pyramids  are  f  " 
as  follows. _ J 

I  now  establish  the  following  list  of  measurements  of 
the  plan  in  connected  natural  numbers.   (See  Figure  4.) 


SOLUTION    OF  THE   PYRAMID    PROBLEM. 


Plan  Ratios  connected  into  Natural  Numbers. 


BY        i  )        48 
[48 
YZ         i  )        48 

BC        6  )        96 

r6 

FB        5  )        80 

DC      45  )      J35 
BC      32)        96 

FB        5  )        80 

r6 

BY        3  )        48 

DN     61  )      183 

t  3 
NR     60  )      180 

DN     61  )      183 

f  3 
NZ      48  )      144 

CY        3  )      144 
U8 
BC        2  )        96 

FH      96  )        96 

DH     55  r    55 

CY      16  )      144 

I  9 
DC      15  )      135 

PN     61  )   146-4 
[2-4 
PA      48  )   115-2 

JE         3)        72 
M4 

EX        2  )        48 

YX        7  )        63 

f  9 

AY        4  )        36 

BA      21  )        84 

(  4 
FB       20  )        80 

CA        4  )      1  80 

MS 

DC        3  )      i35 

BC      32  )        96 

f  3 

EB      21  )        63 

EA        7  }      105 

r  fs 

AZ        4  )        60 

CB*       8  )        96 

(  I2 
BA        7  )        84 

YZ        4  )        48 

(  T2 
AY        3  )        36 

FO      32  )      128 

r4 

OR     21  )        84 

AB        7  )        84 

Vl2 

BO        4  )        48 

ED       8)      120 

MS 

AE        7  )      105 

BA        4  )        84 

r  2I 

EB        3  )        63 

FT      84  )        84 

ST      55  I"'    55 

BC        8  }        96 
Via 
AC      15  )      1  80 

VW    55)       55 
FW     48  )        48 

GE        4  )        96 

r24 

DG       3  )        72 

VW     5S)       55 

M 
SV      36  )        36 

ND     61  )      183 

f  3 
NO     32  )        96 

SJ          7  )        84 

V  12 

SU        6)        72 

HN       4)      128 

[32 
FH        3  )        96 

BJ       45  )      135 
f  3 
AB      28  )        84 

PA   •  48  )   115-2 

f  2'4 
AZ      25  )        60 

The  above  connected  natural  numbers  multiplied  by  eight  become 
R.B.  cubits.  R.B.C. 

(Thus,  BY,  48  x  8  =  384). 


SOLUTION   OF  THE   PYRAMID   PROBLEM. 


Plan  Ratio  Table.— (Continued.} 


GX       2  )      144 

V;2 
DG        i)      '72 

GU       5)      180 
36 
DG       2  )        72 

EO      37)      ii'i 

(  3 
AY      12)        36 

SR  61  )  183 

Y3 

RZ  12  )  36 

SU             2  )            72 

[36 

SV         i  )        36 

HW  144  )      144 

DH     55  i"1    55 

HT     36  )      1  80 

DH     ii  (5    55 

FH  96  )  96 

h 

FE  17)  17 

TW     36)        36 

TU     17)  T     17 

FO        8)      128 
[16 
OZ        3  )        48 

DA      15)      225 

r15 

AZ        4  J        60 

EA  105  }  105 
EF  17)  17 

SR      61  )      183 

[3 

RO     28  )        84 

JB       45  )      i35 

f3 

BY      16)        48 

AC      15)      1  80 

i  I2 

CN        4  )        48 

WH  144  )  144 

HG  17  }'  17 

YW     20  )        80 

(  4 
AY        9  )        36 

FW     48  )        48 
FE      17  I'     17 

YV      15)      135 
AY        4)        36 

TH  180)  '180 
HG  17  f1  17 

•MY       9  )      1  08 

i  I2 
ZY        4  )        48 

AC      20  }      1  80 

\9 

CG        7  )       63 

VZ      61  )      183 

h3 
ZO      id)       48 

AC        9  )      1  80 
[20 
CH       4  )        80 

EA      35)      105 

(3 
AY      12)        36 

EU     84  }        84 
FE      17)        17 

NZ      12)      144 

(  I2 
ZA        5  )        60 

CY        3  )      144 
[48 
YZ         i  )        48 

CA        5  )      1  80 

[36 
AY        i  )       36 

The  above  connected  natural  numbers  multiplied  by  eight  become 
R.B.  cubits.  R.B.C. 

(Thus,  GX  144  x  8  —  1152). 


SOLUTION    OF   THE.  PYRAMID    PROBLEM.  15 


§  2.     THE    ORIGINAL   CUBIT  MEASURE  OF  THE  GIZEH- 

GROUP. 

Mr.  J.'  J.  Wild,  in  his  letter  to  Lord  Brougham  writ- 
ten in  1850,  called  the  base  of  Cephren  seven  seconds. 
I  estimate  the  base  of  Cephren  to  be  just  seven  thir- 
tieths of  the  line  DA.  The  line  DA  is  therefore  thirty 
seconds  of  the  Earth's  Polar  circumference.  The  line 
DA  is  therefore  3033*118625  British  feet,  and  the  base 
of  Cephren  707727  British  feet. 

I  applied  a  variety  of  Cubits  but  found  none  to  work 
in  without  fractions  on  the  beautiful  set  of  natural  di- 
mensions which  I  had  worked  out  for  my  plan.  (See 
table  of  connected  natural  numbers. ) 

I  ultimately  arrived  at  a  cubit  as  the  ancient  measure 
which  I  have  called  the  R.  B.  cubit,  because  it  closely 
resembles  the  Royal  Babylonian  Cubit  of  '5131  metre, 
or  i  '683399  British  feet.  The  difference  is  gir,  of  a  foot. 

I  arrived  at  the  R.  B.  cubit  in  the  following  manner. 

Taking  the  polar  axis  of  the  earth  at  five  hundred  mil- 
lion geometric  inches,  thirty  seconds  of  the  circumference 
will  be  36361-02608 — geometric  inches,  or  36397*4235 
British  inches,  at  nine  hundred  and  ninety-nine  to  the 
thousand — and  3030*0855  geometric  feet,  or  3033*1 18625 
British  feet.  Now  dividing  a  second  into  sixty  parts,  there 
are  1800  R.B.  cubits  in  the  line  DA  ;  and  the  line  DA 
being  thirty  seconds,  measures  36397*4235  British  inches, 
which  divided  by  1800  makes  one  of  my  cubits  20*220- 
7908  British  inches,  or  1*685066  British  feet.  Similarly, 


i6 


SOLUTION   OF   THE   PYRAMID   PROBLEM. 


36361*02608  geometric  inches  divided  by  1800  makes 
my  cubit  20*20057  geometric  inches  in  length.  I  have 
therefore  defined  this  cubit  as  follows: — One  R.B. 
cubit  is  equal  to  20*2006  geo.  inches,  20*2208  Brit, 
inches,  and  i  '685  Brit.  feet. 

I  now  construct  the  following  table  of  measures. 


R.  B.  CUBITS. 

PLETHRA  OR 
SECONDS. 

STADIA. 

MINUTES. 

DEGREES. 

60 

I 

360 

6 

I 

3600 

60 

10 

I 

2l6oOO 

3600 

600 

60 

I 

77760000 

1296000 

2  I6OOO 

2I6OO 

360 

Thus  there  are  seventy-seven  million,  seven  hundred 
and  sixty  thousand  R.B.  cubits,  or  two  hundred  and  six- 
teen thousand  stadia,  to  the  Polar  circumference  of  the 
earth. 

Thus  we  have  obtained  a  perfect  set  of  natural  and 
convenient  measures  which  fits  the  plan,  and  fits  the 
circumference  of  the  earth. 

And  I  claim  for  the  R.B.  cubit  that  it  is  the  most 
perfect  ancient  measure  yet  discovered,  being  the  meas- 
ure of  the  plan  of  the  Pyramids  of  Gizeh. 

The  same  forgotten  wisdom  which  divided  the  cir- 
cle into  three  hundred  and  sixty  degrees,  the  degree 


SOLUTION   OF   THE   PYRAMID    PROBLEM.  I/ 

into  sixty  minutes,  and  the  minute  into  sixty  seconds, 
subdivided  those  seconds,  for  earth  measurements,  into 
the  sixty  parts  represented  by  sixty  R.B.  cubits. 

We  are  aWare  that  thirds  and  fourths  were  used  in 
ancient  astronomical  calculations. 


The  reader  will  now  observe  that  the  cubit  meas- 
ures of  the  main  Pythagorean  triangle  of  the  plan  are 
obtained  by  multiplying  the  original  3,  4  and  5  by  360 ; 
and  that  the  entire  dimensions  are  obtained  in  R.B.  cu- 
bits by  multiplying  the  last  column  of  connected  natural 
numbers  in  the  table  by  eight, — thus— 

R.    B.  CUBITS. 

DC         3  x     360  ='1080 

CA         4  x     360  =  1440 

DA         5  x     360  =  1800 

or, 

NATURAL    NUMBERS.  R.   B.   CUBITS. 

DC  136  x  8  =  1080 
CA  1 80  x  8  =  1440 
DA  225  x  8  =  1800 

&c.,  &c. 
(See  Figure  5,/.  18.) 

According  to  Cassini,  a  degree  was  600  stadia,  a 
minute  10  stadia;  and  a  modern  Italian  mile,  in  the 
year  1723,  was  equal  to  one  and  a  quarter  ancient  Ro- 
man miles  ;  and  one  and  a  quarter  ancient  Roman  miles 
were  equal  to  ten  stadia  or  one  minute.  (Cassini,  Traite 


i8 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


/ 

fig  5. 

J 

J 

/ 

^ 

N 

Ns 

H.B  Cubits. 

FB  =  640     AY=  288 
BA  =    672       ZY=   384 
AF=    S28       ZA=  -380 

J                CA=  1440 
AD=  /800 

DC  »    576 
0               CS  =     768 

/ 

Y 

B 

L 

p 

/,) 

\ 

\ 

C 

EB  =   504 

J3  A                /I  79 

/ 

H 

C 

AE  =   840 
JW  =    768 
OJP  =  J024 

TJV  =  J280 

DZ 

7?.B.  CTMJb. 

/It  level  of  Cepbren's  ^  (Cheops'  Base   420 
Base  ivhich  is  WiepZaneUJ  Cephreifs    do      4^0 
or  level  of  the  plan  J  \_Mycennus'  do       218 

de  la  grandeur  et  de  la  Figure  de  la  Terre.  Amster- 
dam, 1723.) 

Dufeu  also  made  a  stadium  the  six  hundredth  part 
of  a  degree.  He  made  the  degree  110827*68  metres, 
which  multiplied  by  3*280841  gives  363607*996+  British 
feet  ;  and  363607*996+  divided  by  600  equals  606*- 
013327  feet  to  his  stadium. 

I  m&ke  the  stadium  606*62376  British  feet. 

There  being  360  cubits  to  a  stadium,  Dufeu's  stadium 
divided  by  360,  gives  1*6833  British  feet,  which  is  the 
exact  measure  given  for  a  Royal  Babylonian  Cubit,  if 
reduced  to  metres,  viz.  :  0*5131  of  a  metre,  and  therefore 
probably  the  origin  of  the  measure  called  the  Royal 
Babylonian  cubit.  According  to  this  measure,  the  G'i'zeh 
plan  would  be  about  —  smaller  than  if  measured  by 

IOII  J 

R.B.  cubits. 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  19 

§  3.     THE   EXACT   MEASURE   OF   THE   BASES   OF   THE 

PYRAMIDS. 

A  stadium  being  360  R.B.  cubits,  or  six  seconds— 
and  a  plethron  60  R.B.  cubits,  or  one  second,  the  base 
of  the  Pyramid  Cephren  is  seven  plethra,  or  a  stadium 
and  a  plethron,  equal  to  seven  seconds,  or  four  hundred 
and  twenty  R.B.  cubits. 

Mycerinus'  base  is  acknowledged  to  be  half  the  base 
of  Cephren. 

Piazzi  Smyth  makes  the  base  of  the  Pyramid  Cheops 
9131*05  pyramid  (or  geometric)  inches,  which  divided 
by  20*2006  gives  452*01  R.B.  cubits.  I  call  it  452 
cubits,  and  accept  it  as  the  measure  which  exactly  fits 
the  plan. 

I  have  not  sufficient  data  to  determine  the  exact  base 
of  the  other  and  smaller  pyramid  which  I  have  marked 
on  my  plan. 

The  bases,  then,  of  Mycerinus,  Cephren,  and  Cheops, 
are  210,  420  and  452  cubits,  respectively. 

But  in  plan  the  bases  should  be  reduced  to  one  level. 
,  I  have  therefore  drawn  my  plan,  or  horizontal  section, 
at  the  level  or  plane  of  the  base  of  Cephren,  at  which 
level  or  plane  the  bases  or  horizontal  sections  of  the 
pyramids  are — Mycerinus,  218  cubits,  Cephren,  420 
cubits,  and  Cheops,  420  cubits.  I  shall  show  how  I 
arrive  at  this  by-and-by,  and  shall  also  show  that  the 
horizontal  section  of  Cheops,  corresponding  to  the  hori- 
zontal section  of  Cephren  at  the  level  of  Cephren's  base, 


20  SOLUTION  OF  THE   PYRAMID   PROBLEM. 

occurs,  as  it  should  do,  at  the  level  of  one  of  the  courses 
of  masonry,  viz. — the  top  of  the  tenth  course. 

§  4-  THE  SLOPES,  RATIOS,  AND  ANGLES  OF  THE 
THREE  PRINCIPAL  PYRAMIDS  OF  THE  GIZEH 
GROUP. 

Before  entering  on  the  description  of  the  exact  slopes 
and  angles  of  the  three  principal  pyramids,  I  must 
premise  that  I  was  guided  to  my  conclusions  by  making 
full  use  of  the  combined  evolutions  of  the  two  wonderful 
right-angled  triangles,  3,  4,  5,  and  20,  21,  29,  which  seem 
to  run  through  the  whole  design  as  a  sort  of  dominant. 

From  the  first  I  was  firmly  convinced  that  in  such 
skilful  workmanship  some  very  simple  and  easily  applied 
templates  must  have  been  employed,  and  so  it  turned 
out.  Builders  do  not  mark  a  dimension  on  a  plan  which 
they  cannot  measure,  nor  have  a  hidden  measure  of  any 
importance  without  some  clear  outer  way  of  establish- 
ing it. 

This  made  me  "go  straight"  for  the  slant  ratios. 
When  the  Pyramids  were  cased  from  top  to  bottom 
with  polished  marble,  there  were  only  two  feasible 
measures,  the  bases  and  the  apothems  ;*  and  for  that 
reason  I  conjectured  that  these  would  be  the  definite 
plan  ratios. 

Figures  6,  7  and  8  show  the  exact  slope  ratios  of 
Cheops,  Cephren,  and  Mycerinus,  measured  as  shown 

*  The  "  Apothem  is  a  perpendicular  from  the  vertex  of  a  pyramid  on  a  side  of 
the  base."— Chambers'  Practical  Mathematics,  p.  156. 


SOLUTION    OF   THE    PYRAMID    PROBLEM.  21 

on  the  diagrams — viz.,   Cheops,   21  to  34,   Cephren,  20 

Fig.  6. 


Cheops. 


The  .Ratios  of  Bases 
~b  .//Itifcurfes  are  very 
nearly  as  follows,riZ:- 

>ops          33  to  21  or  330  ft  210 
•en      32lv2l  or  320  ft  210 
JVtjternms   .32  to  20  or  336  io  210 


to  33,  and  Mycerinus,   20  to  32 — that  is,   half  base  to 
apothem. 

The  ratios  of  base  to  altitude  are,  Cheops,  33  to  21, 
Cephren,  32  to  21,  and  Mycerinus,  32  to  20  :  not  exactly, 
but  near  enough  for  all  practical  purposes.  For  the 
sake  of  comparison,  it  will  be  well  to  call  these  ratios 
330  to  210,  320  to  210,  and  336  to  210,  respectively. 


Figures  9   and    10  are  meridional  and  diagonal  sec- 


Cheops. 


22 


SOLUTION   OF   THE    PYRAMID    PROBLEM. 


tions,  showing  ratios  of  Cheops,  viz.,  half  base  to 
apothem,  21  to  34  exactly ;  half  base  to  altitude,  53^ 
to  7  nearly,  and  183  to  233,  nearer  still  (being  the  ratio 
of  Piazzi  Smyth).  The  ratio  of  Sir  F.  James,  half  diag- 
onal 10  to  altitude  9  is  also  very  nearly  correct. 

My  altitude  for  Cheops  is  484*887  British  feet,  and 
the  half  base  380.81  British  feet. 

The  ratio  of  7  to  5^  gives  484*66,  and  the  ratio  of 
233  to  183  gives  484*85  for  the  altitude. 

My  half  diagonal  is  538*5465,  and  ratio  10  to  9, 
gives  484*69  British  feet  for  the  altitude. 

I  have  mentioned  the  above  to  show  how  very  nearly 
these  ratios  agree  with  my  exact  ratio  of  21  to  34  half 
base  to  apothem. 


Figures  n  and  12  show  the  ratios  of  Cephren,  viz., 


Jty.ll 


r»en. 


.  12. 

Cephren 


431- 


half  base  to  apothem,  20  to  33  exactly,  and  half  base, 
altitude,  and  apothem  respectively,  as  80,  105,  and  132, 
very  nearly. 

Also  half  diagonal,  altitude,  and  edge,  practically  as 
431,  400,  and  588. 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


Figures  13  and  14  show  the  ratios  of  Mycerinus,  viz., 
Fig.  13.  Fi|M 

JVTycerinus . 


half  base  to  apothem,  20  to  32  exactly,  and  half  base, 
altitude,  and  apothem  respectively,  as  20,  25,  and  32 
very  nearly. 

Also  full  diagonal  to  edge  as  297  to  198,  nearly.  A 
peculiarity  of  this  pyramid  is,  that  base  is  to  altitude  as 
apothem  is  to  half  base,  Thus,  40  :  25  :  :  32  :  20 ;  that 
is,  half  base  is  a  fourth  proportional  to  base,  apothem. 
and  altitude. 

§   5.       THE   EXACT    DIMENSIONS   OF    THE    PYRAMIDS. 
Figures  15  to  20  inclusive,  show  the  linear  dimensions 


Fit  IS. 


Cheops . 


JS.S.Cnb 

452  *  761-62 

287-76?  -484-887 
365  9047.  616-549 
430058  -72^647 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


of  the  three  pyramids,  also  their  angles.  The  base 
angles  are,  Cheops,  51°  51'  20";  Cephren,  52°  41'  41"; 
and  Mycerinus,  51°  19'  4". 


Fig.  IT. 


210 


Cephren . 


Ceph 


re/i. 


Tig.  20. 


Br&Ft. 

420  =•  707-  70 
27561^46440 
34650  =  58385 
40516=68265 
59397=100084 


Cub,  Brit  Ft: 
210          =353-85 
168  =28308 

JT31-J4  s^20v97 
198-20  =333?9S 
296-9848*50042 


SOLUTION   OF  THE   PYRAMID    PROBLEM.  2$ 

In  Cheops,  my  dimensions  agree  with  Piazzi  Smyth 
—in  the  base  of  Cephren,  with  Vyse  and  Perring — in 
the  height  of  Cephren,  with  Sir  Gardner  Wilkinson, 
nearly — in  the  base  of  Mycerinus,  they  agree  with  the 
usually  accepted  measures,  and  in  the  height  of  Myce- 
rinus, they  exceed  Jas.  J.^Vild's  measure,  by  not  quite 
one  of  my  cubits. 

In  my  angles  I  agree  very  nearly  with  Piazzi  Smyth, 
for  Cheops,  and  with  Agnew,  for  Cephren,  differing 
about  half  a  degree  from  Agnew,  for  Mycerinus,  who 
took  this  pyramid  to  represent  the  same  relation  of  n 
that  P.  Smyth  ascribes  to  Cheops  (viz.  :  51°  51'  i4*"3), 
while  he  gave  Cheops  about  the  same  angle  which  I  as- 
cribe to  Mycerinus. 

I  shall  now  show  how  I  make  Cephren  and  Cheops 
of  equal  bases  of  420  R.B.  cubits  at  the  same  level,, 
viz. — that  of  Cephren's  base. 

John  James  Wild  made  the  bases  of  Cheops,  Cephren, 
and  Mycerinus,  respectively,  80,  100,  and  104*90  cubits 
above  some  point  that  he  called  Nile  Level. 

His  cubit  was,  I  believe,  the  Memphis,  or  Nilometric 
cubit — but  at  any  rate,  he  made  the  base  of  Cephren 
412  of  them. 

I  therefore  divided  the  recognized  base  of  Cephren 
—viz.,  70775  British  feet — by  412,  and  got  a  result  of 
17178  British  feet  for  his  cubit.     Therefore,  his  meas- 
ures multiplied  by   17178    and  divided    by   1*685  will 
turn  his  cubits  into  R.B.  cubits. 

I  thus  make  Cheops,  Cephren,   and  Mycerinus,  re- 


26  SOLUTION   OF  THE  PYRAMID    PROBLEM. 

spectively,  81*56,  101*93,  and  106*93  R.B.  cubits  above 
the  datum  that  J.  J.  Wild  calls  Nile  Level.  According 
to  Bonwick's  "  Facts  and  Fancies,"  p.  31,  high  water 
Nile  would  be  138^  ft.  below  base  of  Cheops  (or  82.19 
R.B.  cubits). 

Piazzi  Smyth  makes  the  pavement  of  Cheops  1752 
British  inches  (or  86*64  R.B.  cubits)  above  average 
Nile  Level,  but,  by  scaling  his  map,  his  high  Nile  Level 
appears  to  agree  nearly  with  Wild. 

It  is  the  relative  levels  of  the  Pyramids,  however^ 
that  I  require,  no  matter  how  much  above  Nile 
Level. 

Cephren's  base  of  420  cubits  being  101*93  cubits, 
and  Cheops'  base  of  452  cubits  being  81*56  cubits  above 
Wild's  datum,  the  difference  in  level  of  their  bases 
is,  20*37  cubits. 

The  ratio  of  base  to  altitude  of  Cheops  being  330 
to  210,  therefore  20*37  cubits  divided  by  210  and  mul- 
tiplied by  330  equals  32  cubits;  and  452  cubits  minus 
32  cubits,  equals  420. 

Similarly,  the  base  of  Mycerinus  is  5  cubits  above 
the  base  of  Cephren,  and  the  ratio  of  base  to  altitude 
32  to  20  ;  therefore,  5  cubits  divided  by  20  and  multi- 
plied by  32  equals  8  cubits  to  be  added  to  the  210 
cubit  base  of  Mycerinus,  making  it  218  cubits  in 
breadth  at  the  level  of  Cephren's  base. 

Thus,  a  horizontal  section  or  plan  at  the  level  of 
Cephren's  base  would  meet  the  slopes  of  the  Pyramids 
so  that  they  would  on  plan  appear  as  squares  with  sides 


SOLUTION   OF   THE   PYRAMID    PROBLEM.  2/ 

equal  to  218.  420,  and  420  R.B.  cubits,  for  Mycerinus, 
Cephren,  and  Cheops,  respectively. 

Piazzi  Smyth  makes  the  top  of  the  tenth  course  of 
Cheops  414  pyramid  inches  above  the  pavement ;  and 
414  divided  by  20*2006  equals  20*49  R-B.  cubits. 

But  I  have  already  proved  that  Cheops'  420  cubit 
base  measure  occurs  at  a  level  of  20*37  cubits  above 
pavement ;  therefore  is  this  level  the  level  of  the  top  of 
the  tenth  course,  for  the  difference  is  only  0*12  R.B. 
cubits,  or  2^2  inches. 


I  wish  here  to  note  as  a  matter  of  interest,  but  not 
as  affecting  my  theory,  the  following  measures  of  Piazzi 
Smyth,  turned  into  R.B.  cubits,  viz.  :— 


PYR.    INCHES.  R.B.    CUBITS. 


King's  Chamber  floor,  above  pavement . .  1702'         —    84/25 

Cheops'  Base,  as  before  stated. 913 1*05     =  452*01 

King's  Chamber,  "  True  Length,". ......    412*132  —     20^40 

"           "True  First  Height,".   230-389  =     ir4o 
"  True  Breadth," 2o6'o66  =     10*20 

He  makes  the  present  summit  platform  of  Cheops 
5445  pyramid  inches  above  pavement.  My  calculation 
of  269*80  .  R.B.  cub.  (See  Fig.  21)  is  equal  to  5450 
pyramid  inches — this  is  about  18  cubits  below  the 
theoretical  apex. 


Figure   21   represents  the    comparative   levels   and 
dimensions  of  Mycerinus,  Cephren,  and  Cheops. 


28  SOLUTION   OF   THE   PYRAMID    PROBLEM. 

The  following  peculiarities  are  noticeable  : — That 
Cheops  and  Cephren  are  of  equal  bases  at  the  level  of 
Cephren's  base  ; — that,  at  the  level  of  Cheops'  base,  the 
latter  is  only  half  a  cubit  larger ; — that,  from  the  level 
of  Mycerinus'  base,  Cheops  is  just  double  the  height 
of  Mycerinus  ; — and  that  from  the  level  of  Cephren's 
base,  Cephren  is  just  double  the  height  of  Mycerinus  ; 
measuring  in  the  latter  case,  however,  only  up  to  the 
level  platform  at  the  summit  of  Cephren,  which  is  said 
to  be  about  ei^ht  feet  wide. 

o 

The  present  summit  of  Cephren  is  23*07  cubits 
above  the  present  summit  of  Cheops,  and  the  com- 
pleted apex  of  Cephren  would  be  8*21  cubits  above  the 
completed  apex  of  Cheops. 

In  the  summit  platforms  I  have  been  guided  by 
P.  Smyth's  estimate  of  height  deficient,  363  pyr.  inches, 
for  Cheops,  and  I  have  taken  8  feet  base  for  Cephren's 
summit  platform. 

§    6.     GEOMETRICAL   PECULIARITIES    OF    THE 
PYRAMIDS. 

In  any  pyramid,  the  apothcm  is  to  half  the  base  as 
the  area  of  the  four  sides  is  to  the  area  of  the  base. 
Thus — Ratio  apothem  to  half  base  Mycerinus. 3 2  to  20 

"     Cephren.. ..33  to  20 
"     Cheops 34  to  2 1 

AREA   OF  THE    FOUR   SIDES.  AREA    OF  THE    BASE. 

Mycerinus 70560'          44100 

Cephren 291060*          176400 

Cheops 33°777'9°     204304 

All  in  R.B.  cubits. 


SOLUTION   OF   THE   PYRAMID    PROBLEM.  29 

Therefore  —  32   :   20    :  :    70560'   :    44100 

33  :   20    :  :   291060'   :   176400 

34  :   21    :  :   33°777'9°  :   204304 

*  Herodotus  states  that  "  the  area  of  each  of  the  four 
faces  of  Cheops  was  equal  to  the  area  of  a  square 
whose  base  was  the  altitude  of  a  Pyramid  ;  "  or,  in 
other  words,  that  altitude  was  a  mean  proportional  to 
apothem  and  half  base  ;  thus  —  :area  of  one  face  equals 
the  fourth  of  330777*90  or  82694*475  R.B.  cubits,  and  the 
square  root  of  82694*475  is  287*56.  But  the  correct 
altitude  is  287*77,  so  tne  error  is  0*21,  or  4^  British 
inches.  I  have  therefore  the  authority  of  Herodotus  to 
support  the  theory  which  I  shall  subsequently  set  forth, 
that  this  pyramid  was  the  exponent  of  lines  divided  in 
mean  and  extreme  ratio. 

By  taking  the  dimensions  of  the  Pyramid  from  what 
I  may  call  its  working  level,  that  is,  the  level  of  the  base 
of  Cephren,  this  peculiarity  shows  more  clearly,  as  also 
others  to  which  I  shall  refer.  Thus  —  base  of  Cheops  at 
working  level,  420  cubits,  and  apothem  340  cubits  ; 
base  area  is,  therefore,  i  76400  cubits,  and  area  of  one 
face  is  (420  cubits,  multiplied  by  half  apothem,  or  170 
cubits)  71400  cubits.  Now  the  square  root  of  71400 
would  give  altitude,  or  side  of  square  equal  to  altitude, 
267*207784  cubits  :  but  the  real  altitude  is  A/34O2  —  2io2 
=  ^71500  =  267*394839.  So  that  the  error  of  Hero- 
dotus's  proposition  is  the  difference  between  ^714  and 


"Proctor  is  responsible  for  this  statement,  as  I  am  quoting  from  an  essay  of 
his  in  the  Gentleman  s  Magazine.      R.  B. 


30  SOLUTION   OF   THE   PYRAMID   PROBLEM. 

This  leads  to  a  consideration  of  the  properties  of 
the  angle  formed  by  the  ratio  apothcm  34  to  half  base 
21,  peculiar  to  the  pyramid  Cheops.  (See  Figiire  22.) 


t/Jrea  of  Square 
with  side  equal 
to  Secant 


Fig.  22 


Diagram  ifluslrakn 
relations  of  ratios  ^ 
the  pyramid  CHEOPS. 


Calling  apothem  34,  radius ;    and  half  base  21,  sine 
—I  find  that— 

Radius  is  the  square  root  of  1156 

Sine 441 

Co-sine 715 

Tangent 713 

Secant 1869 

and  Co-versed-sine..  . .  , 169 

So  it  follows  that  the  area  of  one  of  the  faces,  714,  is  a 
mean  between  the  square  of  the  altitude  or  co-sine,  715, 
and  the  square  of  the  tangent,  713. 

Thus  the  reader  will  notice  that  the  peculiarities  of 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  31 

the  Pyramid  Cheops  lie  in  the  regular  relations  of  the 
squares  of  its  various  lines  ;  while  the  peculiarities  of 
the  other  two  pyramids  lie  in  the  relations  of  the  lines 
themselves. 

Mycerinus  and  Cephren,  born,  as  one  may  say,  of 
those  two  noble  triangles  3,  4,  5,  and  20,  21,  29,  exhibit 
in  their  lineal  developments  ratios  so  nearly  perfect 
that,  for  all  practical  purposes,  they  may  be  called 
correct. 

Thus  —  Mycerinus,  (a)  2O2  +  ^S2—  1025,  and  322=  1024. 
and  Cephren,  (b)  8o2  +  io52  —  1  7425,  and  1  322  = 

17424. 
or  (c)  4002  +  43  1  2^  345  761,  and 


See  diagrams,  Figures  11  to  14  inclusive. 

In  the  Pyramid  Cheops,  altitude  is  very  nearly  a 
mean  proportional  between  apothem  and  half  base. 
Apothem  being  34,  and  half  base  21,  then  altitude 
would  be  A/342—  2i2  ^  ^715  =  267394839,  and— 

21    :  267394839   :   :  267394839  :  34,  nearly. 

Here,  of  course,  the  same  difference  comes  in  as  oc- 
curred in  considering  the  assumption  of  Herodotus,  viz., 
the  difference  between  ^715  and  A/7  14  ;  because  if  the 
altitude  were  A/714,  then  would  it  be  exactly  a  mean 
proportional  between  the  half  base  and  the  apothem  ; 
(thus,  21  :  2672077  :  :  2672077  :  :  34.) 


(a)  Half  base  to  altitude,     (b)  Half  base   to  altitude,     (c)  Half    diagonal  of 
base  to  altitude. 


32  SOLUTION   OF   THE   PYRAMID    PROBLEM. 

In  Cheops,  the  ratios  of  apothem,  half  base  and 
edge  are,  34,  21,  and  40,  very  nearly,  thus,  342  x  2i2— 
1597,  and  40* =  1600. 

The   dimensions  of  Cheops  (from  the  level  of  the 
base  of  Cephren)  to  be  what    Piazzi  Smyth   calls  a  * 
pyramid,  would  be- 
Half  base  210  R.B.  cubits. 
Altitude     267-380304,  &c. 
Apothem  339*988573,  &c. 

Altitude  being  to  perimeter  of  base,  as  radius  of  a  circle 
to  circumference. 

My  dimensions  of  the  pyramid  therefore  in  which — 

Half  base  =  210  R.B.  cubits. 

Altitude     =  267-394839  &c. 
Apothem  =  340  " 

come  about  as  near  to  the  ratio  of  K  as  it  is  possible  to 
come,  and  provide  simple  lines  and  templates  to  the 
workmen  in  constructing  the  building ;  and  I  entertain 
no  doubt  that  on  the  simple  lines  and  templates  that 
my  ratios  provide,  were  these  three  pyramids  built. 

§  6A  THE  CASING  STONES  OF  THE  PYRAMIDS. 

Figures  23,  24,  and  2  5,  represent  ordinary  casing  stones 
of  the  three  pyramids,  and  Figures  26,  27,  and  28, 
represent  angle  or  quoin  casing  stones  of  the  same. 

The  casing  stone  of  Cheops,  found  by  Colonel 
Vyse,  is  represented  in  Bonwiek's  "  Pyramid  Facts  and 
Fancies,"  page  16,  as  measuring  four  feet  three  inches  at 
the  top,  eight  feet  three  inches  at  the  bottom,  four  feet 


Eg.  23. 


Casing  Stones 


Eg.  26 


•Angle  or  Quoin 


-. 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


33 


eleven  inches   at  the  back,  and  six  feet  three  inches  at 
the  front.      Taking  four  feet  eleven   inches  as  Radius^ 
and  six  feet  three  inches  as  Secant,  then  the  Tangent  is 
three  feet  ten  inches  and  three  tenths. 
Thus,  in  inches 


2  —  59")  =  46*30  inches  ;  there- 
fore the  inclination  of  the  stone  must  have  been — slant 
height  75  inches  to  46*30  inches  horizontal.  Now^ 
46*30  is  to  75,  as  21  is  to  34.  Therefore,  Col.  Vyse's 
casing  stone  agrees  exactly  with  my  ratio  for  the  Pyramid 
Cheops,  viz.,  21  to  34.  (See  Figiire  29.) 


~  v,,   .     ,  j  o^o 

Casing    £tone. 


6ft.  3in 


75: 46-3'.:  33:  21 


This  stone  must  have  been  out  of  plumb  at  the  back 
an  inch  and  seven  tenths  ;  perhaps  to  give  room  for 
grouting  the  back  joint  of  the  marble  casing  stone  to 
the  limestone  body  of  the  work :  or,  because,  as  it  is 
not  a  necessity  in  good  masonry  that  the  back  of  a 
stone  should  be  exactly  plumb,  so  long  as  the  error  is 


34 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


on  the  right  side,  the  builders  might  not  have  been  par- 
ticular in  that  respect. 

Figure  59  represents  such  a  template  as  the  masons 


l  of  Cheops,  slanding  'crt  angle  ofwall.) 


would  have  used  in  building  Cheops,  both  for  dressing 
and  setting  the  stones.  (The  courses  are  drawn  out  of 
proportion  to  the  template.)  The  other  pyramids  must 
have  been  built  by  the  aid  of  similar  templates. 

Such  large  blocks  of  stone  as  were  used  in  the  casing 
of  these  pyramids  could  not  have  been  completely 
dressed  before  setting  ;  the  back  and  ends,  and  the  top 
and  bottom  beds  were  probably  dressed  off  truly,  and 
the  face  roughly  scabbled  off  ;  but  the  true  slope  angle 
could  not  have  been  dressed  off  until  the  stone  had  been 
truly  set  and  bedded,  otherwise  there  would  have  been 
great  danger  to  the  sharp  arrises. 


I  shall  now  record  the  peculiarities  of  the  3,  4,  5  or 


Fig.  30. 


Fig.  31. 


22789^ 


QD 

3360 


8690 


Fig.  35. 


SOLUTION  OF  THE  PYRAMID  PROBLEM.         35 

Pythagorean  triangle,  and  the   right-angled  triangle  20, 
21,  29. 

§  7.     PECULIARITIES   OF  THE  TRIANGLES  3,  4,  5,  AND 

20,    21,    29. 

The  3,  4,  5    triangle  contains  36°  52'  11*65"  and  the 
complement  or  greater  angle  53°  7'  48*35" 

Radius 5  60  whole  numbers.* 

Co-sine.  • 4  48 

Sine 3  36 

Versed  sine  . . .  i  12  " 

Co-versed  sine    2  24 

Tangent 33/4  45 

Secant 6ti    =        75 

Co-tangent 62A    =       80 

Co-secant 85^    =      100 

Tangent   +    Secant  =   Diameter  or  2  Radius 
Co  tan.  +   Co-sec    =   3  Radius 
Sine     :    Versed-sine     1:3:1 
Co-sine     :    Co-versed  sine     1:2:1 

Figure  30  illustrates  the  preceding  description. 
Figure  31  shows  the  3*1  triangle,  and  the  2*1  triangle 
built  up  on  the  sine  and  co-sine  of  the  3,  4,  5  triangle. 
The  3*1  triangle  contains  18°  26'  5*82"  and  the  2*1 
triangle  26°  33'  54*1 9'' ;  the  latter  has  been  frequently 
noticed  as  a  pyramid  angle  in  the  gallery  inclinations. 

Figure  32  shows  these  two  triangles  combined  with 
the  3,  4,  5  triangle,  on  the  circumference  of  a  circle, 

*  Co  =  3X4x5 


36         SOLUTION  OF  THE  PYRAMID  PROBLEM. 

The  20,  21,  29  triangle  contains  43°  36'  10*15"  and  the 
complement,  46°  23'  49-85". 
Expressed  in  whole  numbers — 

Radius 29     =      12180* 

Sine 20     =        8400 

Co-sine 21      —        8820 

Versed  sine...   8     =        3360 
Co-versed  sine  9     =        3  780 

Tangent =      1 1 600 

Co-tangent....          =      12789 

Secant =      16820 

Co-sec =      17661 

Tangent   +    Secant  =   2l/i  radius 
Co-tan.    +   Co-sec  =   2/^  radius 
Sine     :    Versed  sine     1:5:2 
Co-sine     :    Co-versed  sine    :  :    7     :    3 

It  is  noticeable  that  while  the  multiplier  required  to 
bring  radius  5  and  the  rest  into  whole  numbers,  for  the 
3,  4,  5  triangle  is  twelve,  in  the  20,  21,  29  triangle  it  is 
420,  the  key  measure  for  the  bases  of  the  two  main  pyr- 
amids in  R.B.  cubits. f 

I  am  led  to  believe  from  study  of  the  plan,  and  con- 
sideration of  the  whole  numbers  in  this  20,  21,  29  tri- 
angle, that  the  R.B.  cubit,  like  the  Memphis  cubit,  was 
divided  into  280  parts. 

The  whole  numbers  of  radius,  sine,  and  co-sine  divided 
by  280,  give  a  very  pretty  measure  and  series  in  R.B. 
cubits,  viz.,  434,  30,  and  3 1 J,  or  87, 60,  and  63,  or  1 74,  1 20 

*  12180  =  20  X  21    x    29  f  12  —  3  X4,and  420  =  20  x  21 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


37 


and  126; — all  exceedingly  useful  in  right-angled  meas- 
urements. Notice  that  the  right-angled  triangle  174, 
1 20,  126,  in  the  sum  of  its  sides  amounts  to  420. 

Figure  33  illustrates  the  20,  21,  29  triangle.  Figure 
34  shows  the'5'2  and  7*3  triangles  built  up  on  the  sine 
and  co-sine  of  the  20,  21,  29  triangle. 

The  5  '2  triangle  contains  21°  48'  5*08"  and  the  7*3 
triangle  23°  n'  54-98". 

Figure  35  shows  how  these  two  triangles  are  combined 
with  the  20,  21,  29  triangle  on  the  circumference,  and 
Figure  36  gives  a  general  view  and  identification  of 


Fig.  36. 

Halios  of 
L  eading  Triangles . 


these  six  triangles  which  occupied  an  important  position 
in  the  trigonometry  of  a  people  who  did  all  their  work 
by  right  angles  and  proportional  lines. 


3§  SOLUTION   OF   THE   PYRAMID    PROBLEM. 

§  8.     GENERAL    OBSERVATIONS. 

It  must  be  admitted  that  in  the  details  of  the  build- 
ing of  the  Pyramids  of  G'izeh  there  are  traces  of  other 
measures  than  R.B.  cubits,  but  that  the  original  cubit 
of  the  plan  was  1*685  British  feet  I  feel  no  doubt.  It 
is  a  perfect  and  beautiful  measure,  fit  for  such  a  noble 
design,  and,  representing  as  it  does  the  sixtieth  part  of 
a  second  of  the  Earth's  polar  circumference,  it  is  and 
was  a  measure  for  all  time. 

It  may  be  objected  that  these  ancient  geometricians 
could  not  have  been  aware  of  the  measure  of  the 
Earth's  circumference  ;  and  wisely  so,  were  it  not  for 
two  distinct  answers  that  arise.  The  first  being,  that 
since  I  think  I  have  shown  that  Pythagoras  never  dis- 
covered the  Pythagorean  triangle,  but  that  it  must  have 
been  known  and  practically  employed  thousands  of  years 
before  his  era,  in  the  Egyptian  Colleges  where  he 
obtained  his  M.A.  degree,  so  in  the  same  way  it  is 
probable  that  Eratosthenes,  when  he  went  to  work  to 
prove  that  the  earth's  circumference  was  fifty  times  the 
distance  from  Syene  to  Alexandria,  may  have  obtained 
the  idea  from  his  ready  access  to  the  ill-fated  Alexan- 
drian Libfary,  in  which  perhaps  some  record  of  the 
learning  of  the  builders  of  the  Pyramids  was  stored. 
And  therefore  I  claim  that  there  is  no  reason  why  the 
pyramid  builders  should  not  have  known  as  much  about 
the  circumference  of  the  earth  as  the  modern  world  that 
has  calmly  stood  by  in  its  ignorance  and  permitted 


SOLUTION   OF   THE    PYRAMID   PROBLEM.  39 

those  magnificent  and,  as  I  shall  prove,  useful  edifices 
to  be  stripped  of  their  beautiful  garments  of  polished 
marble. 

My  second  answer  is  that  the  correct  cubit  measure 
may  have  been  got  by  its  inventors  in  a  variety  of  other 
ways ;  for  instance,  by  observations  of  shadows  of 
heavenly  bodies,  without  any  knowledge  even  that  the 
earth  was  round  ;  or  it  may  have  been  evolved  like  the 
British  inch,  which  Sir  John  Herschel  tells  us  is  within 
a  thousandth  part  of  being  one  five  hundred  millionth 
of  the  earth's  polar  axis.  I  cloubt  if  the  circumference 
of  the  earth  was  considered  by  the  inventor  of  the 
British  inch. 

It  was  a  peculiarity  of  the  Hindoo  mathematicians 
that  they  tried  to  make  out  that  all  they  knew  was  very 
old.  Modern  savants  appear  to  take  the  opposite  stand 
for  any  little  information  they  happen  to  possess. 

The  cubit  which  is  called  the  Royal  Babylonian 
cubit  and  stated  to  measure  0*5131  metre,  differs  so 
slightly  from  my  cubit,  only  the  six-hundredth  part  of  a 
foot,  that  it  may  fairly  be  said  to  be  the  same  cubit,  and  it 
will  be  for  antiquaries  to  trace  the  connection,  as  this 
may  throw  some  light  on  the  identity  of  the  builders  of 
the  Pyramids  of  Gizeh.  Few  good  English  two-foot 
rules  agree  better  than  these  two  cubits  do. 

While  I  was  groping  about  in  the  dark  searching 
for  this  bright  needle,  I  tried  on  the  plan  many  likely 
ancient  measures. 

For  a  long  time  I  worked  in  Memphis  or  Nilometric 


40  SOLUTION   OF   THE   PYRAMID   PROBLEM. 

cubits,  which  I  made  17126  British  feet;  they  seem  to 
vary  from  170  to  172,  and  although  I  made  good  use 
of  them  in  identifying  other  people's  measures,  still  they 
were  evidently  not  in  accordance  with  the  design  ;  but 
the  R.B.  cubit  of  i;685  British  feet  works  as  truly  into 
the  plan  of  the  Pyramids  without  fractions  as  it  does 
into  the  circumference  of  the  earth. 

Here  I  might,  to  prevent  others  from  falling  into 
one  of  my  errors,  point  out  a  rock  on  which  I  was 
aground  for  a  long  time.  I  took  the  base  of  the  Pyra- 
mid Cheops,  determined  by  Piazzi  Smyth,  from  Bonwick's 
"Pyramid  Facts  and  Fancies"  (a  valuable  little  refer- 
ence book),  as  763.81  British  feet,  and  the  altitude  as 
486.2567;  and  then  from  Piazzi  Smyth's  "  Inheritance," 
page  27,  I  confirmed  these  figures,  and  so  worked  on 
them  for  a  long  time,  but  found  always  a  great  flaw  in 
my  work,  and  at  last  adopted  a  fresh  base  for  Cheops, 
feeling  sure  that  Mr.  Smyth's  base  was  wrong  :  for  I  was 
absolutely  grounded  in  my  conviction  that  at  a  certain 
level,  Cheops'  and  Cephren's  measures  bore  certain  rela- 
tions to  each  other.  I  subsequently  found  in  another 
part  of  Mr.  Smyth's  book,  that  the  correct  measures 
were  761.65  and  484.91  British  feet  for  base  and  altitude, 
which  were  exactly  what  I  wanted,  and  enabled  me  to 
be  in  accordance  with  him  in  that  pyramid  which  he 
appears  to  have  made  his  particular  study. 

For  the  information  of  those  who  may  wish  to  com- 
pare my  measures,  which  are  the  results  of  an  even  or 
regular  circumference  without  fractions,  with  Mr.  Smyth's 


.SOLUTION   OF   THE   PYRAMID    PROBLEM.  41 

measures,  which  are  the  results  of  an  even  or  regular 
diameter  without  fractions,  it  may  be  well  to  state  that 
there  are  just  about  99  R.B.  cubits  in  80  of  Piazzi 
Smyth's  cubits  of -25  pyramid  inches  each. 

§  9.     THE    PYRAMIDS    OF    EGYPT,    THE    THEODOLITES. 
OF    THE    EGYPTIAN    LAND    SURVEYORS. 

About  twenty-three  years  ago,  on  my  road  to  Aus- 
tralia, I  was  crossing  from  Alexandria  to  Cairo,  and  saw 
the  pyramids  of  Gi'zeh. 

I  watched  them  carefully  as  the  train  passed  along, 
noticed  their  clear  cut  lines  against  the  sky,  and  their 
constantly  changing  relative  position. 

I  then  felt  a  strong  conviction  that  they  were  built 
for  at  least  one  useful  purpose,  and  that  purpose  was 
the  survey  of  the  country.  I  said,  "  Here  be  the  Theod- 
olites of  the  Egyptians." 

Built  by  scientific  men,  well  versed  in  geometry,  but 
unacquainted  with  the  use  of  glass  lenses,  these  great 
stone  monuments  are  so  suited  in  shape  for  the  purposes 
of  land  surveying,  that  the  practical  engineer  or  sur- 
veyor must,  after  consideration,  admit  that  they  may 
have  been  built  mainly  for  that  purpose. 

Not  only  might  the  country  have  been  surveyed  by 
these  great  instruments,  and  the  land  allotted  at  period- 
ical times  to  the  people  ;  but  they,  remaining  always  in 
one  position,  were  there  to  correct  and  readjust  bound- 
aries destroyed  or  confused  by  the  annual  inundations  of 
the  Nile. 


42  SOLUTION   OF   THE    PYRAMID   PROBLEM. 

The  Pyramids  of  Egypt  may  be  considered  as  a 
great  system  of  landmarks  for  the  establishment  and 
easy  readjustment  at  any  time  of  the  boundaries  of  the 
holdings  of  the  people. 

The  Pyramids  of  Gizeh  appear  to  have  been 
main  marks ;  and  those  of  Abousir,  Sakkarah,  Dashow, 
Lisht,  Meydoun,  &c.,  with  the  great  pyramids  in  Lake 
Mceris,  subordinate  marks,  in  this  system,  which  was 
probably  extended  from  Chaldea  through  Egypt  into 
Ethiopia. 

The  pyramid  builders  may  perhaps  have  made  the 
entombment  of  their  Kings  one  of  their  exoteric  objects, 
playing  on  the  morbid  vanity  of  their  rulers  to  induce 
them  to  the  work,  but  in  the  minds  of  the  builders 
before  ever  they  built  must  have  been  planted  the  inten- 
tion to  make  use  of  the  structures  for  the  purposes  of 
land  surveying. 

The  land  of  Egypt  was  valuable  and  maintained  a 
dense  population  ;  every  year  it  was  mostly  submerged, 
and  the  boundaries  destroyed  or  confused.  Every 
soldier  had  six  to  twelve  acres  of  land  ;  the  priests  had 
their  slice  of  the  land  too  ;  after  every  war  a  reallot- 
ment  of  the  lands  must  have  taken  place,  perhaps  every 
year. 

While  the  water  was  lying  on  the  land,  it  so  softened 
the  ground  that  the  stone  boundary  marks  must  have 
required  frequent  readjustment,  as  they  would  have  been 
likely  to  fall  on  one  side. 

By  the  aid  of  their  great  stone  theodolites,  the  sur- 


SOLUTION   OF   THE   PYRAMID   PROBLEM.  43 

veyors,  who  belonged  to  the  priestly  order,  were  able 
to  readjust  the  boundaries  with  great  precision.  That 
all  science  was  comprised  in  their  secret  mysteries  may 
be  one  reason  why  no  hieroglyphic  record  of  the  scien- 
tific uses  of  the  pyramids  remains.  It  is  possible  that  at 
the  time  of  Diodorus  and  Herodotus,  (and  even  when 
Pythagoras  visited  Egypt,)  theology  may  have  so 
smothered  science,  that  the  uses  of  the  pyramids  may 
have  been  forgotten  by  the  very  priests  to  whom  in 
former  times  the  knowledge  belonged  ;  but  "  a  respectful 
reticence  "  which  has  been  noticed  in  some  of  these  old 
writers  on  pyramid  and  other  priestly  matters  would 
rather  lead  us  to  believe  that  an  initiation  into  the  mys- 
teries may  have  sealed  their  lips  on  subjects  about  which 
they  might  otherwise  have  been  more  explicit. 

The  ''closing"  of  one  pyramid  over  another  in 
bringing  any  of  their  many  lines  into  true  order,  must 
even  now  be  very  perfect; — but  now  we  can  only 
imagine  the  beauties  of  these  great  instrumental  won- 
ders of  the  world  when  the  casing  stones  were  on  them. 
We  can  picture  the  rosy  lights  of  one,  and  the  bright 
white  lights  of  others ;  their  clear  cut  lines  against  the 
sky,  true  as  the  hairs  of  a  theodolite ;  and  the  sombre 
darkness  of  the  contrasting  shades,  bringing  out  their 
angles  with  startling  distinctness.  Under  the  influence 
of  the  Eastern  sun,  the  faces  must  have  been  a  very 
blaze  of  light,  and  could  have  been  seen  at  enormous 
distances  like  great  mirrors. 

I   declare  that   the  pyramids  of  G'fzeh  in  all  their 


44  SOLUTION    OF   THE   PYRAMID   PROBLEM. 

polished  glory,  before  the  destroyer  stripped  them 
of  their  beautiful  garments,  were  in  every  respect 
adapted  to  flash  around  clearly  defined  lines  of  sight, 
upon  which  the  lands  of  the  nation  could  be  accurately 
threaded.  The  very  thought  of  these  mighty  theodo- 
lites of  the  old  Egyptians  fills  me  with  wonder  and 
reverence.  What  perfect  and  beautiful  instruments 
they  were  !  never  out  of  adjustment,  always  correct, 
always  ready  ;  no  magnetic  deviation  to  allow  for.  No 
wonder  they  took  the  trouble  they  did  to  build  them  so 
correctly  in  their  so  marvellously  suitable  positions. 

If  Astronomers  agree  that  observations  of  a 
pole  star  could  have  been  accurately  made  by  peering 
up  a  small  gallery  on  the  north  side  of  one  of  the  pyra- 
mids only  a  few  hundred  feet  in  length,  I  feel  that 
I  shall  have  little  difficulty  in  satisfying  them  that 
accurate  measurements  to  points  only  miles  away  could 
have  been  made  from  angular  observations  of  the  whole 
group. 

§    io.     HOW   THE    PYRAMIDS   WERE    MADE    USE    OF. 

It  appears  from  what  I  have  already  set  forth  that 
the  plan  of  the  Pyramids  under  consideration  is 
geometrically  exact,  a  perfect  set  of  measures. 

I  shall  now  show  how  these  edifices  were  applied  to 
a  thoroughly  geometrical  purpose  in  the  true  meaning 
of  the  word — to  measure  the  Earth. 

I  shall  show  how  true  straight  lines  could  be 
extended  from  the  Pyramids  in  given  directions  useful 


SOLUTION   OF  THE    PYRAMID   PROBLEM.  45 

in  right-angled  trigonometry,  by  direct  observation 
of  the  buildings,  and  without  the  aid  of  other  instru- 
ments. 

And  I  shall  show  how  by  the  aid  of  a  simple  instru- 
ment angles  could  be  exactly  observed  from  any 
point. 

This  Survey  theory  does  not  stand  or  fall  on  the 
merits  of  my  theory  of  the  G'izeh  plan.  Let  it  be 
proved  that  this  group  is  not  built  on  the  exact  system 
of  triangulation  set  forth  by  me,  it  is  still  a  fact  that  its 
plan  is  in  a  similar  shape,  and  any  such  shape  would 
enable  a  surveyor  acquainted  with  the  plan  to  lay  down 
accurate  surveys  by  observations  of  the  group  even 
should  it  not  occupy  the  precise  lines  assumed  by  me. 

And    here    I    must    state    that    although    the    lines 

o 

of  the  plan  as  laid  down  herein  agree  nearly  with  the 
lines  as  laid  down  in  Piazzi  Smyth's  book,  in  the 
Penny  Cyclopaedia,  and  in  an  essay  of  Proctor's  in  the 
Gentleman  s  Magazine,  still  I  find  that  they  do  not  agree 
at  all  satisfactorily  with  a  map  of  the  Pyramids  in 
Sharp's  "  Egypt,"  said  to  be  copied  from  Wilkinson's 
map. 

We  will,  however,  for  the  time,  and  to  explain  my 
survey  theory,  suppose  the  plan  theory  to  be  correct,  as 
I  firmly  believe  it  is. 

And  then,  supposing  it  may  be  proved  that  the 
respective  positions  of  the  pyramids  are  slightly  dif- 
ferent to  those  that  I  have  allotted  to  them  on  my 
plan,  it  will  only  make  a  similar  slight  difference  to 


46  SOLUTION   OF  THE   PYRAMID   PROBLEM. 

the  lines  and  angles  which  I  shall  here  show  could  be 
laid  out  by  their  aid. 

Let  us  in  the  first  place  comprehend  clearly  the 
shape  of  the  land  of  Egypt. 

A  sector  or  fan,  with  a  long  handle — the  fan  or 
sector,  the  delta  ;  and  the  handle  of  the  fan,  the  Nile 
Valley,  running  nearly  due  south. 

The  Pyramids  of  Gi'zeh  are  situate  at  the  angle  of 
the  sector,  on  a  rocky  eminence  whence  they  can  all  be 
seen  for  many  miles.  The  summits  of  the  two  high 
ones  can  be  seen  from  the  delta,  and  from  the  Nile 
Valley  to  a  very  great  distance  ;  how  far,  I  am  unable 
to  say ;  but  I  should  think  that  while  the  group  could 
be  made  general  use  of  for  a  radius  of  fifteen  miles,  the 
summits  of  Cephren  and  Cheops  could  be  made  use  of 
for  a  distance  of  thirty  miles  ;  taking  into  consideration 
the  general  fall  of  the  country. 

It  must  be  admitted  that  if  meridian  observations  of 
the  star  Alpha  of  the  Dragon  could  be  made  with 
accuracy  by  peeping  up  a  small  hole  in  one  of  the 
pyramids,  then  surely  might  the  surveyors  have  carried 
true  north  and  south  lines  up  the  Nile  Valley  as  far  as 
the  summit  of  Cheops  was  visible,  by  "plumbing  zn" 
the  star  and  the  apex  of  the  pyramid  by  the  aid  of 
a  string  and  a  stone. 

True  east  and  west  lines  could  have  been  made  to 
intersect  such  north  and  south  lines  from  the  various 
groups  of  pyramids  along  the  river  banks,  by  whose  aid 
also  such  lines  would  be  prolonged. 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  47 

Next,  supposing  that  their  astronomers  had  been 
aware  of  the  latitude  of  Cheops,  and  the  annual  north- 
ing and  southing  of  the  sun,  straight  lines  could  have 
been  laid  out  in  various  sectoral  directions  to  the  north- 
eastward and  north-westward  of  Cheops,  across  the 
delta,  as  far  as  the  extreme  apex  of  the  pyramid  was 
visible,  by  observations  of  the  sun,  rising  or  setting 
over  his  summit.  (That  the  Dog-star  was  observed  in 
this  manner  from  the  north-west,  I  have  little  doubt.) 

For  this  purpose,  surveyors  would  be  stationed  at 
suitable  distances  apart  with  their  strings  and  their 
stones,  ready  to  catch  the  sun  simultaneously,  and  at 
the  very  moment  he  became  transfixed  upon  the  apex 
of  the  pyramid,  and  was,  as  it  were,  "swallowed  by  it." 
(See  Figure  37.)  The  knowledge  of  the  pyramid  slope 


i   37. 


angle  from  different  points  of  view  would  enable  the 


48  SOLUTION   OF   THE    PYRAMID   PROBLEM. 

surveyor  to  place  himself  in  readiness  nearly  on  the 
line. 

Surely  such  lines  as  these  would  be  as  true  and  as 
perfect  as  we  could  lay  out  nowadays  with  all  our 
modern  instrumental  appliances.  A  string1  and  a  stone 
here,  a  clean-cut  point  of  stone  twenty  miles  away,  and 
a  great  ball  of  fire  behind  that  point  at  a  distance  of 
ninety  odd  million  miles.  The  error  in  such  a  line 
would  be  very  trifling. 

Such  observations  as  last  mentioned  would  have 
been  probably  expended  from  Cephren  for  long  lines,  as 
being  the  higher  pyramid  above  the  earth's  surface,  and 
may  have  been  made  from  the  moon  or  stars. 

In  those  days  was  the  sun  the  intimate  friend  of 
man.  The  moon  and  stars  were  his  hand-maidens. 

How  many  of  us  can  point  to  the  spot  of  the  sun's 
rising  or  setting  ?  We,  with  our  clocks,  and  our 
watches,  and  our  compasses,  rarely  observe  the  sun  or 
stars.  But  in  a  land  and  an  age  when  the  sun  was  the 
only  clock,  and  the  pyramid  the  only  compass,  the 
movements  and  positions  of  the  heavenly  bodies  were 
known  to  all.  These  people  were  familiar  with  the 
stars,  and  kept  a  watch  upon  their  movements. 

How  many  of  our  vaunted  educated  population 
could  point  out  the  Dog-star  in  the  heavens  ? — but  the 
whole  Egyptian  nation  hailed  his  rising  as  the  begin- 
ning of  their  year,  and  as  the  harbinger  of  their  annual 
blessing,  the  rising  of  the  waters  of  the  Nile. 

It  is  possible  therefore  that  the  land  surveyors  of 


!5         S1 

fcto-SJ*    **!** 

*  g  2  Co    §§i£? 


SOLUTION   OF   THE   PYRAMID    PROBLEM.  49 

Egypt  made  full  use  of  the  heavenly  bodies  in  their 
surveys  of  the  land  ;  and  while  we  are  pitifully  laying 
out  our  new  countries  by  the  circumferenter  and  the 
compass,  we  presume  to  speak  slightingly  of  the  sup- 
posed dark  heathen  days,  when  the  land  of  Egypt  was 
surveyed  by  means  of  the  sun  and  the  stars,  and  the 
theodilites  were  built  of  stone,  with  vertical  limbs  five 
hundred  feet  in  height,  and  horizontal  limbs  three  thou- 
sand feet  in  diameter. 

Imagine  half  a  dozen  such  instruments  as  this  in  a 
distance  of  about  sixty  miles  (for  each  group  of  pyra- 
mids was  effectually  such  an  instrument),  and  we  can 
form  some  conception  of  the  perfection  of  the  surveys 
of  an  almost  prehistoric  nation. 

The  centre  of  Lake  Moeris,  in  which  Herodotus  tells 
us  two  pyramids  stood  300  feet  above  the  level  of  the 
lake,  appears  from  the  maps  to  be  about  S.  28°  W.,  or  S. 
29°  W.  from  G'izeh,  distant  about  57  miles,  and  the  Mei- 
dan  group  of  pyramids  appears  to  be  about  33  miles 
due  south  of  G'izeh. 

Figures  38,  39,  40  and  41,  show  that  north-west, 
south-east,  north-east,  and  south-west  lines  from  the 
pyramids  could  be  extended  by  simply  plumbing  the 
angles.  These  lines  would  be  run  in  sets  of  two's  and 
three's,  according  to  the  number  of  pyramids  in  the 
group ;  and  their  known  distances  apart  at  that  angle 
would  check  the  correctness  of  the  work. 

A  splendid  line  was  the  line  bearing  43°  36'  10*15", 
or  223°  36'  io'i5"  from  Cheops  and  Cephren,  the  pyra- 


50  SOLUTION   OF  THE   PYRAMID   PROBLEM. 

mids  covering  each  other,  the  line  of  hypotenuse  of  the 
great  20,  21,  29  triangle  of  the  plan.  This  I  call  the 
20,  21  line.  (See  Figiire  42.) 

Figure  43  represents  the  3,  4,  5  triangle  line  from 
the  summits  of  Mycerinus  and  Cheops  in  true  line  bear- 
ing 216°  52'  1 1  "65".  This  I  call  the  south  4,  west  3  line. 

The  next  line  is  what  I  call  the  2,  i  line,  and  is 
illustrated  by  figure  44.  It  is  one  of  the  most  perfect 
of  the  series,  and  bears  S.  26°  33'  54*19"  W.  from  the 
apex  of  Cephren.  This  line  demonstrates  clearly  why 
Mycerinus  was  cased  with  red  granite. 

Not  in  memory  of  the  beautiful  and  rosy-cheeked 
Nitocris,  as  some  of  the  tomb  theory  people  say,  but  for 
a  less  romantic  but  more  useful  object  ;  simply  because, 
from  this  quarter,  and  round  about,  the  lines  of  the 
pyramids  would  have  been  confused  if  Mycerinus  had 
not  been  of  a  different  color.  The  2,  i  line  is  a  line  in 
which  Mycerinus  would  have  been  absolutely  lost  in 
the  slopes  of  Cephren  but  for  his  red  color.  There  is 
not  a  fact  that  more  clearly  establishes  my  theory,  and 
the  wisdom  and  forethought  of  those  who  planned  the 
G'fzeh  pyramids,  than  this  red  pyramid  Mycerinus,  and 
the  2,  i  line. 

Hekeyan  Bey,  speaks  of  this  pyramid  as  of  a  "  ruddy 
complexion;"  John  Greaves  quotes  from  the  Arabic 
book,  Morat  Alzeman,  "  and  the  lesser  which  is  col- 
ourcd ;  "  and  an  Arabic  writer  who  dates  the  Pyramids 
three  hundred  years  before  the  Flood,  and  cannot  find 
among  the  learned  men  of  Egypt  "  any  certain  relation 


Fig.  42 


21  West  20. 
Bearing  223^36'.  10-157 


$oulh  4.  Wesl  3. 
Bearing  216*52'.  11  65" 


Fig. 

South  2. 

Bearing  206^33:53-18? 


Fig.  45. 

Dearing  209^48:32  8J 

Fig.  46. 

Soulh  3,  Westl. 
Bearing  198 1 26:5*2 


Beari7ig20r.4$'.5: 


Fig.  IB. 

SovXh  7  West  3. 
Bearing  SOSlll'.SS: 


SOLUTION   OF   THE   PYRAMID    PROBLEM.  51 

concerning  them"  nor  any  "  memory  of  them  amongst 
men"  also  expatiates  upon  the  beauties  of  the  "coloured 
satin  "  covering  of  this  one  particular  pyramid. 

Figure  45  represents  the  line  south  96,  west  55, 
from  Cephren,.  bearing  209°  48'  32*81";  the  apex  of 
Cephren  is  immediately  above  the  apex  of  Mycerinus. 

Figure  46  is  the  S.  3  W.  i  line,  bearing  198°  26'  5.82" ; 
here  the  dark  slope  angle  of  the  pyramids  with  the  sun 
to  the  eastward  occupies  half  of  the  apparent  half 
base. 

Figure  47  is  the  S.  5,  W.  2  line,  bearing  201°  48'  5"  ; 
here  Cephren  and  Mycerinus  are  in  outside  slope  line. 

Figure  48  is  the  S.  7  W.  3  line,  bearing  203°  n'  55''; 
here  the  inside  slope  of  Cephren  springs  from  the 
centre  of  the  apparent  base  of  Mycerinus. 

I  must  content  myself  with  the  preceding  examples 
of  a  few  pyramid  lines,  but  must  have  said  enough  to 
show  that  from  every  point  of  the  compass  their  ap- 
pearance was  distinctly  marked  and  definitely  to  be 
determined  by  surveyors  acquainted  with  the  plan. 

§11.     DESCRIPTION   OF   THE   ANCIENT    PORTABLE 
SURVEY   INSTRUMENT. 

I  must  now  commence  with  a  single  pyramid,  show 
how  approximate  observations  could  be  made  from  it, 
and  then  extend  the  theory  to  a  group  with  the  obser- 
vations thereby  rendered  more  perfect  and  delicate. 

We  will  suppose  the  surveyor  to  be  standing  look- 
ing at  the  pyramid  Cephren  ;  he  knows  that  its  base  is 


52  SOLUTION   OF   THE    PYRAMID    PROBLEM. 

420  cubits,  and  its  apothem  346^  cubits.  He  has  provided 
himself  with  a  model  in  wood,  or  stone,  or  metal,  and 
one  thousandth  of  its  size — therefore  his  model  will  be 
0.42  cubit  base,  and  0.3465  cubit  apothem  —  or,  in 
round  numbers,  eight  and  half  inches  base,  and  seven 
inches  apothem. 

This  model  is  fixed  on  the  centre  of  a  card  or 
disc,  graduated  from  the  centre  to  the  circumference, 
like  a  compass  card,  to  the  various  points  of  the  com- 
pass, or  divisions  of  a  circle. 

The  model  pyramid  is  fastened  due  north  and  south 
on  the  lines  of  this  card  or  disc,  so  that  when  the  north 
point  of  the  card  points  north,  the  north  face  of  the 
model  pyramid  faces  to  the  north. 

The  surveyor  also  has  a  table,  which,  with  a  pair  of 
plumb  lines  or  mason's  levels,  he  can  erect  quite  level : 
this  table  is  also  graduated  from  the  centre  with  divis- 
ions of  a  circle,  or  points  of  the  compass,  and  it  is  larger 
than  the  card  or  disc  attached  to  the  model. 

This  table  is  made  so  that  it  can  revolve  upon  its 
stand,  and  can  be  clamped.  We  will  call  it  the  lower 
limb.  There  is  a  pin  in  the  centre  of  the  lower  limb, 
and  a  hole  in  the  centre  of  the  disc  bearing  the  model, 
which  can  be  thus  placed  upon  the  centre  of  the  table, 
and  becomes  the  upper  limb.  The  upper  limb  can  be 
clamped  to  the  lower  limb. 

The  first  process  will  be  to  clamp  both  upper  and 
lower  limbs  together,  with  the  north  and  south  lines  of 
both  in  unison,  then  revolve  both  limbs  on  the  stand  till 


SOLUTION   OF   THE   PYRAMID   PROBLEM.  53 

the  north  and  south  line  points  straight  for  the  pyramid 
in  the  distance,  which  is  done  by  the  aid  of  sights 
erected  at  the  north  and  south  points  of  the  perimeter 
of  the  lower  limb.  When  this  is  adjusted,  clamp  the 
lower  limb  and  release  the  upper  limb  ;  now  revolve  the 
upper  limb  until  the  model  pyramid  exactly  covers  the 
pyramid  in  the  distance,  and  shows  just  the  same  shade 
on  one  side  and  light  on  the  other,  when  viewed  from 
the  sights  of  the  clamped  lower  limb  —  and  the  lines, 
angles,  and  shades  of  the  model  coincide  with  the  lines, 
angles,  and  shades  of  the  pyramid  observed ; — now 
clamp  the  upper  limb.  Now  does  the  model  stand 
really  due  north  and  south,  the  same  as  the  pyramid  in 
the  distance  ;  it  throws  the  same  shades,  and  exhibits 
the  same  angles  when  seen  from  the  same  point  of 
view  ;  just  as  much  of  it  is  in  shade  and  as  much  of  it 
is  in  light  as  the  pyramid  under  observation  ;  therefore 
it  must  be  standing  due  north  and  south,  because  Ceph- 
ren  himself  is  standing  due  north  and  south,  and  the 
upper  limb  reads  off  on  the  lower  limb  the  angle  or 
bearing  observed. 

So  far  we  possess  an  instrument  equal  to  the  modern 
circumferenter,  and  yet  we  have  only  brought  one  pyra- 
mid into  work. 

If  I  have  shown  that  such  an  operation  as  the  above 
is  practically  feasible,  if  I  have  shown  that  angles  can 
be  taken  with  moderate  accuracy  by  observing  one 
pyramid  of  420  cubits  base,  how  much  more  accurate 
will  the  observation  be  when  the  surveyor's  plane  table 


54  SOLUTION   OF  THE   PYRAMID   PROBLEM. 

bears  a  group  of  pyramids  which  occupy  a  representative 
space  of  about  1400  cubits  when  viewed  from  the  south 
or  north,  and  about  1 760  cubits  when  viewed  from  the 
east  or  west.  If  situated  a  mile  or  two  south  of  the 
G'fzeh  group  our  surveyor  could  also  tie  in  and  perfect 
his  work  by  sights  to  the  Sakkarah  group  with  Sakkarah 
models  ;  and  so  on,  up  the  Nile  Valley,  he  would  find 
every  few  miles  groups  of  pyramids  by  aid  of  which  he 
would  be  enabled  to  tie  his  work  together. 

If  the  G'fzeh  group  of  pyramids  is  placed  and  shaped 
in  the  manner  I  have  described,  it  must  be  clear  that  an 
exact  model  and  plan,  say  a  thousandth  of  the  size,  could 
be  very  easily  made — the  plan  being  at  the  level  of  the 
base  of  Cephren  where  the  bases  of  the  two  main 
pyramids  are  even  ; — and  if  they  are  not  exactly  so 
placed  and  shaped,  it  may  be  admitted  that  their  position 
and  dimensions  were  known  to  the  surveyors  or  priests, 
so  that  such  models  could  be  constructed.  It  is  probable, 
therefore,  that  the  instrument  used  in  conjunction  with 
these  pyramids,  was  a  machine  constructed  in  a  similar 
manner  to  the  simple  machine  I  have  described,  only 
instead  of  there  being  but  one  model  pyramid  on  the 
disc  or  upper  limb,  it  bore  the  whole  group  ;  and  the 
smaller  pyramids  were  what  we  may  call  vernier  points 
in  this  great  circle,  enabling,  the  surveyor  to  mark  off 
known  angles  with  great  'accuracy  by  noticing  how,  as 
he  worked  round  the  group  of  pyramids,  one  or  other 
of  the  smaller  ones  was  covered  by  its  neighbours.* 

*  See  general  plan  of  Gizeh  Group  op.  page  I. 


SOLUTION   OF   THE   PYRAMID    PROBLEM.  $5 

The  immensity  of  the  main  pyramids  would  require 
the  smaller  ones  to  be  used  for  surveys  in  the  im- 
mediate neighbourhood,  as  the  surveyor  might  easily 
be  too  close  to  get  accurate  observations  from  the  main 
pyramids. 

The  upper  limb,  then,  was  -a  disc  or  circular  plate 
bearing  the  model  of  the  group. 

Cheops  would  be  situated  in  the  centre  of  the  circle, 
and  observat:ons  would  be  taken  by  bringing  the  whole 
model  group  into  even  line  and  even  light  and  shade 
with  the  GTzeh  group. 

I  believe  that  with  a  reasonable-sized  model  occupy- 
ing a  circle  of  six  or  seven  feet  diameter,  such  as  a 
couple  of  men  could  carry,  very  accurate  bearings  could 
have  been  taken,  and  probably  were  taken. 

The  pyramid  shape  is  the  very  shape  of  all  others 
to  employ  for  such  purposes.  A  cone  would  be  useless, 
because  the  lights  and  shades  would  be  softened  off  and 
its  angles  from  all  points  would  be  the  same.  Other 
solids  with  perpendicular  angles  would  be  useless, 
because  although  they  would  vary  in  width  from  different 
points  of  view  they  would  not  present  that  ever  chang- 
ing angle  that  a  pyramid  does  when  viewed  from  differ- 
ent directions. 

After  familiarity  with  the  models  which  I  have  made 
use  of  in  prosecuting  these  investigations,  I  find  that  I 
can  judge  with  great  accuracy  from  their  appearance 
only  the  bearing  of  the  group  from  any  point  at  which 
I  stand.  I  make  bold  to  say  that  the  pocket  compass 


56  SOLUTION   OF   THE   PYRAMID   PROBLEM. 

of  the  Egyptian  surveyor  was  a  little  model  of  the  group 
of  pyramids  in  his  district,  and  he  had  only  to  hold  it 
up  on  his  hand  and  turn  it  round  in  the  sun  till  its 
shades  and  angles  corresponded  with  the  appearance  of 
the  group,  to  tell  as  well  as  we  could  tell  by  our  com- 
passes, perhaps  better,  his  bearing  from  the  landmarks 
that  governed  his  surveys. 

The  Great  Circle  of  Gold  described  by  Diodorus 
(Diod.  Sic.  lib.  X.,  part  2,  cap.  i)  as  having  been  em- 
ployed by  the  Egyptians,  and  on  which  was  marked 
amongst  other  things,  the  position  of  the  rising  and 
setting  of  the  stars,  and  stated  by  him  to  have  been 
carried  off  by  Cambysses  when  Egypt  was  con- 
quered by  the  Persians,  is  supposed  by  Cassini  to 
have  been  also  employed  for  finding  the  meridian 
by  observation  of  the  rising  and  setting  of  the  sun. 
This  instrument  and  others  described  by  writers 
on  Egypt  would  have  been  in  practice  very  similar 
to  the  instrument  which  I  have  described  as  hav- 
ing been  probably  employed  for  terrestrial  observa- 
tions. 

The  table  or  disc  comprising  the  lower  limb  of  the 
instrument,  might  have  been  supported  upon  a  small 
stand  with  a  circular  hole  in  the  centre,  so  arranged  that 
the  instrument  could  be  either  set  up  alone  and  sup- 
ported by  its  own  tripod,  or  rested  fairly  on  the  top  of 
any  of  those  curious  stone  boundary  marks  which  were 
made  use  of,  not  only  to  mark  the  corners  of  the  differ- 
ent holdings,  but  to  show  the  level  of  the  Nile  inunda- 


SOLUTION    OF    THE   PYRAMID   PROBLEM.  57 

tions.      (See    Figure  49,    copied  from   Sharpes  Egypt, 


vol.  /.,/.  6.)  The  peculiar  shape  of  the  top  of  these 
stone  landmarks,  or  "  sacred  boundary  stones,"  appears 
suitable  for  such  purposes,  and  it  would  have  been  a 
great  convenience  to  the  surveyor,  and  conducive  to 
accuracy,  that  it  should  be  so  arranged  that  the  instru- 
ment should  be  fixed  immediately  over  the  mark,  as 
appears  probable  from  the  shape  of  the  stone. 

A  noticeable  point  in  this  theory  is,  that  it  is  not  in 
the  least  essential  that  the  apex  of  a  pyramid  should  be 
complete.  If  their  summits  were  left  permanently  flat, 
they  would  work  in  for  survey  purposes  quite  as  well, 
and  I  think  better,  than  if  carried  to  a  point,  and  they 
would  be  more  useful  with  a  flat  top  for  defined  shadows 
when  used  as  sun  dials. 

In  the  G'i'zeh  group,  the  summit  of  Cheops  appears 
to  me  to  have  been  left  incomplete  the  better  to  get  the 
range  with  Cephren  for  lines  down  the  delta. 

In  this  system  of  surveying,  there  is  always  a  beau- 
tiful connection  between  the  horizontal  bearings  and  the 
apparent  or  observed  angles  presented  by  the  slopes  and 
edges  of  the  pyramid.  Thus,  in  pyramids  like  those  of 
G'i'zeh,  which  stand  north  and  south,  and  whose  meridi- 
onal sections  contain  less,  and  whose  diagonal  sections 
contain  more  than  a  right  angle,  the  vertex  being  the 
point  at  which  the  angle  is  measured  —  this  law  holds  :  — 


SOLUTION   OF   THE   PYRAMID    PROBLEM. 


That  the  smallest  interior  angle  at  the  vertex,  con- 
tained between  the  inside  edge  and  the  outside 
edge,  will  exhibit  the  same  angle  as  the  bearing 
of  the  observer's  eye  from  the  apex  of  the 
pyramid  when  the  angle  at  the  apex  contained 
by  the  outside  edges  appears  to  be  a  right  angle. 

Figures  50  to  55   inclusive  illustrate   this  beautiful 

Cheops 

IfOin  pouils  bea 
B    19.12.22" 
13.12.22 
J9-J2.22 
19. 12  «22 


Cheops 

from  points  bearing 
$  19 .12,22  W 
W19 .12.22  3 
fT  19. 12.22  IV 
E  19*12,22  Jf 


Cephrcn  -from 

points  TjewiTi 

S  23.7.50"'24 

W 23.  7.  50-24 

Jf  23.  7.  60-24 

E  23-  7"  50-24 


Lev, 

,    Poi 


Celiren  from 
nts  1)  earing 


..  7.  50-24 
23  •  7*  30-24 


v* 


Cheops  model 


Cephren  Model 


Fit.57 


model 


SOLUTION    OF  THE    PYRAMID    PROBLEM.  59 

law,  from  which  it  will  be  seen  that  the  Gi'zeh  surveyors 
Jnycerinus  ^ 

from  points  ~hea-rin£  \'"'      *"v 

fl  f7J.40a.4  W  °  \      -'  -« 

W17.1.4Q-4N 
X17.1.404E 

E 17. 1.40  4  $ 


.-.. 

Fig.  55. 

Jnycerinus  \ 


&  J7.1.404 
W 17- 1.40-4 
JV 17*  1.40-4  W 
E  J7*l~40-4  JV 


/ 
/    %^ 


pO3sessed,  in  this  manner  alone,  eight  distinctly  defined 
bearings  from  each  pyramid. 


I  recommend  any  one  desirous  to  thoroughly  com- 
prehend these  matters,  to  make  a  plan  from  my  diagram, 
Figure  5,  using  R.B.  cubits  for  measures,  and  to  a  suit- 
able scale,  on  a  piece  of  card-board.  Then  to  cut  out 
of  the  card-board  the  squares  of  the  bases  of  the  pyra- 
mids at  the  level  of  Cephren,  viz.,  420,  420  and  218 
cubits  respectively,  for  the  three  main  pyramids.  One 
hundred  cubits  to  the  inch  is  a  convenient  scale  and 
within  the  limits  of  a  sheet  of  Bath  board. 

By  striking  out  the  models  on  card-board  in  the 
manner  shown  by  diagrams  (see  Figures  56,  57,  and  58) 
they  can  be  cut  out  with  a  penknife — cutting  only  half 


60  SOLUTION   OF   THE    PYRAMID    PROBLEM. 

through  where  the  lines  are  dotted — bent  up  together, 
and  pasted  along  the  edges  with  strips  of  writing  paper 
about  half  an  inch  wide. 

These  models  can  be  dropped  into  the  squares  cut 
out  of  the  card-board  plan,  thus  correcting  the  error 
caused  by  the  thickness  of  the  card-board  base,  and  if 
placed  in  the  sun,  or  at  night  by  the  light  of  one  lamp 
or  candle  properly  placed  to  represent  the  sun  in  the 
eastward  or  westward,  the  clear  cut  lines  and  clear  con- 
trasting shades  will  be  manifest,  and  the  lines  illustrated 
by  my  figures  can  be  identified. 

When  inspecting  the  model,  it  is  well  to  bear  in 
mind  that  the  eye  must  be  kept  very  nearly  level  with 
the  table,  or  the  pyramids  will  appear  as  if  viewed  from 
a  balloon. 


I  believe  that  the  stones  were  got  up  to  the  building 
by  way  of  the  north  side  of  each  pyramid.  The  casing 
on  the  south,  east,  and  west,  was  probably  built  up  as 
the  work  proceeded,  and  the  whole  of  these  three  faces 
were  probably  thus  finished  and  completed  while  there 
was  not  a  single  casing  stone  set  on  the  north  side. 
Then  the  work  would  be  closed  up  until  there  remained 
nothing  but  a  great  gap  or  notch,  wide  at  the  bottom, 
and  narrowing  to  the  apex.  The  work  on  the  north 
side  would  then  be  closed  from  the  sides  and  top,  and 
the  bottom  casing  stone  about  the  centre  of  the  north 
side,  would  be  the  last  stone  set  on  the  building. 


SOLUTION   OF   THE   PYRAMID   PROBLEM.  6 1 

These  old  builders  were  too  expert  not  to  have  thus 
made  use  of  all  the  shade  which  their  own  building 
would  thus  afford  to  a  majority  of  the  workmen. 


Many  of  the  obelisks  were  probably  marks  on  pyra- 
mid lines  of  survey. 

The  pyramid  indeed  may  have  been  a  development 
of  the  obelisk  for  this  purpose. 

Their  slanting  sides  might  correspond  with  some  of 
the  nearly  upright  slant  angles  of  the  pyramids,  in  posi- 
tions opposite  certain  lines.  Reference  to  several  of 
my  figures  will  show  how  well  this  would  come  in. 

Herodotus  speaks  of  two  obelisks  at  Heliopolis,  and 
Bonwick  tells  us  that  Abd  al  Latif  saw  two  there  which 
he  called  Pharaoh's  Needles.  An  Arab  traveller,  in 
1190,  saw  a  pyramid  of  copper  on  the  summit  of  the 
one  that  remained,  but  it  is  now  wanting.  Pharaoh's 
Needles  appear  to  have  been  situated  about  20  miles 
NE.  of  the  G'izeh  group,  and  their  slope  angles  might 
have  coincided  with  the  apparent  slope  angles  of  Ce- 
phren  or  Cheops  on  the  edge  nearest  the  obelisk. 

The  ancient  method  of  describing  the  meridian  by 
means  of  the  shadow  of  a  ball  placed  on  the  summit  of 
an  obelisk  points  to  a  reasonable  interpretation  for  the 
peculiar  construction  of  the  two  pillars,  Jachin  and 
Boaz,  which  are  said  to  have  been  situated  in  front  of 
the  Hebrew  Temple  at  Jerusalem,  and  about  which  so 
much  mysterious  speculation  has  occurred. 


62  SOLUTION    OF   THE   PYRAMID    PROBLEM. 

They  were  no  doubt  used  as  sun-dials  for  the  morn- 
ing and  afternoon  sun  by  the  shadow  of  the  balls  or 
\  "  chapiters  "  thrown  upon  the  pavement. 

Without  presuming  to  dispute  the  objects  assigned 
by  others  for  the  galleries  and  passages  which  have  been 
discovered  in  the  pyramid  Cheops,  I  venture  to  opine 
that  they  were  employed  to  carry  water  to  the  builders. 
They  are  connected  with  a  well,  and  the  well  with  the 
Nile  or  canal.  Whether  the  water  was  slided  up  the 
smooth  galleries  in  boxes,  or  whether  the  cochlea,  or 
water  screw,  was  worked  in  them,  their  angles  being 
suitable,  it  is  impossible  to  conjecture  ;  either  plan 
would  have  been  convenient  and  feasible. 

These  singular  chambers  and  passages  may  indeed 
possibly  have  had  to  do  with  some  hydraulic  machinery 
of  great  power  which  modern  science  knows  nothing 
about.  The  section  of  the  pyramid,  showing  these 
galleries,  in  the  pyramid  books,  has  a  most  hydraulic 
appearance. 

The  tremendous  strength  and  regularity  of  the  cavi- 
ties called  the  King's  and  Queen's  chambers,  the  regu- 
larity and  the  smallness  of  most  of  the  passages  or  mas- 
sive stone  connecting  pipes,  favor  the  idea  that  the  cham- 
bers might  have  been  reservoirs,  their  curious  roofs, 
air  chambers,  and  the  galleries  or  passages,  connecting 
pipes  for  working  water  under  pressure.  Water  raised 
through  the  passages  of  this  one  pyramid  nearest  to  the 
canal,  might  have,  been  carried  by  troughs  to  the  other 
pyramids,  which  were  in  all  probability  in  course  of  con- 


SOLUTION   OF   THE    PYRAMID    PROBLEM.  63 

struction  at  the  same  period  of  time.  A  profane  friend 
of  mine  thinks  that  the  sarcophagus  or  "  sacred  coffer" 
in  the  King's  chamber  may  have  been  used  by  the  chief 
architect  and  leading  men  of  the  works  as  a  bath,  and 
that  the  King's  chamber  was 
than  a  delightful  bath  room.  V  °p  THR. 

UNIVERSITY 


The  following  quotation  from  the^vutlllgof  an 
Arabian  author  (Ibn  Abd  Alkokm),  is  extracted  from 
Bonwick's  "  Pyramid  Facts  and  Fancies,"  page  72  :— 
"  The  Coptites  mention  in  their  books  that  upon  them 
(the  Pyramids)  is  an  inscription  engraven  ;  the  expo- 
sition of  it  in  Arabicke  is  this  : — '  I,  Saurid  the  King 
built  the  Pyramids  (in  such  and  such  a  time),  and 
finished  them  in  six  years  ;  he  that  comes  after  me,  and 
says  he  is  equal  to  me,  let  him  destroy  them  in  six 
hundred  years  ;  and  yet  it  is  known  that  it  is  easier  to 
pluck  down  than  to  build  ;  and  when  I  had  finished 
them,  I  covered  them  with  sattin,  and  let  him  cover  them 
with  slats'  ' 

The  italics  are  my  own.  The  builder  seems  to  have 
entertained  the  idea  that  his  work  would  be  partially 
destroyed,  and  afterwards  temporarily  repaired  or 
rebuilt.  The  first  part  has  unfortunately  come  true, 
and  it  is  possible  that  the  last  part  of  the  idea  of  King 
Saurid  may  be  carried  out,  because  it  would  not  be  so 
very  expensive  an  undertaking  for  any  civilized  nation 
in  the  interest  of  science  to  re-case  the  pyramids 


64 


SOLUTION   OF   THE    PYRAMID    PROBLEM. 


of   G'i'zeh,    so  that  they  might  be    once   more    applied 
to  land-surveying  purposes  in  the  ancient  manner. 

It  would  not  be  absolutely  necessary  to  case  the 
whole  of  the  pyramid  faces,  so  long  as  sufficient  casing 
was  put  on  to  define  the  angles.  The  "slats"  used 
might  be  a  light  wooden  framework  covered  with  thin 
metal.  The  metal  should  be  painted  white,  except 
in  the  case  of  Mycerinus,  which  should  be  of  a  reddish 
color. 

Main   Triangular    Dimensions   of   Plan   are    Represented 
by  the  Following  Eight  Right-angled  Triangles. 


TABLE    TO    EXPLAIN   FIGURE    60. 


AB 

28  )   x  ( 

84 

)   x  (    672 

DG     3)   > 

(    r  } 

x   (    576 

BJ 

45  K 

135 

(  8  ]  Io8° 

GE        4       2, 

\    96  [ 

g  ]    768 

JA 

53)       •( 

159 

(1272 

ED     5  \    A 

-    (       120      ) 

(    9^° 

DC 
CA 
AD 

sl4*5! 

135 

1  80 

225 

1  x  (  I08° 
f  «  1  !44o 

)  8  (  1800 

FW   48  ) 
WV  55  [  ^ 
VF    73)    ] 

(          48      ) 

55 
(      73    ) 

x  (    384 
o   •<    44° 
*   1    584 

EB 

3  }        ( 

63 

)   x   (    5°4 

FB     20) 

(      8°    ) 

(        640 

BA 

4  f  X  \ 

84 

-    o    ]     672 

BA    21  V  > 

\       84    C 

Q     ]      672 

AE 

s)"( 

)   8   (    S4o 

AF    29)   L 

(     il6    ) 

(      928 

FH 

3)   x   ( 

96 

)   x  (    I68 

NOTE.  —  In 

the  above  table  the  first 

HN 

4  [   *] 

128- 

r  a  1  I024 

column   is  the  Ratio,   the 

second  the 

NF 

s}32( 

1  60 

)  6  (  1280 

connected  Natural  Numbers,   and  the 

third  column 

represents  // 

le  length  of 

AY 
YZ 

S)xj 

4f  12  ] 

36 

48 

)  x  (    288 
8       3B4 

each  line  in  R 

.B.  cubits. 

ZA 

s\  2( 

60 

)       (    480 

SOLUTION   OF   THE    PYRAMID    PROBLEM. 


12.  PRIMARY  TRIANGLES  AND  THEIR  SATELLITES  ; 
—OR  THE  ANCIENT  SYSTEM  OF  RIGHT-ANGLED 
TRIGONOMETRY  UNFOLDED  BY  A  STUDY  OF  THE 
PLAN  OF  THE  PYRAMIDS  OF  GIZEH. 

Reference  to  Fig.  60  and   the  preceding  table,  will 


show  that  the  main  triangular  dimensions  of  this  plan 
(imperfect  as  it  is  from  the  lack  of  eleven  pyramids) 
are  represented  by  four  main  triangles,  viz  :— 

Ratio. 
C  A  D  C  . .  3,     4,     5 

*J  '  I    "  V-J 

F  B  A  F  . .      . .   20,  21,  29 

A  B  J  A  . .      . .    28,  45,  53 
FWVF  .,      ..   48,  55,  73 

Figures  30  to  36  illustrate  the  two  former,  and  Figures 
6 1  and  62  illustrate  the  two  latter.  I  will  call  triangles 
of  this  class  "  primary  triangles,"  as  the  most  suitable 
term,  although  it  is  applied  to  the  main  triangles  of 
geodetic  surveys. 


66  SOLUTION   OF    THE    PYRAMID    PROBLEM. 

We  have  only  to  select  a  number  of  such  triangles 
and  a  system  of  trigonometry  ensues,  in  which  base, 
perpendicular,  and  hypotenuse  of  every  triangle  is  a 
whole  measure  without  fractions,  and  in  which  the 
nomenclature  for  every  angle  is  clear  and  simple. 

An  angle  of  43°  36'  io'i5"  will  be  called  a  20,  21 
angle,  and  an  angle  of  36°  52'  11*65"  w'1^  be  called  a 
3,  4  angle,  and  so  on. 

In  the  existing  system  whole  angles,  such  as  40,  45, 
or  50  degrees,  are  surrounded  by  lines,  most  of  which 
can  only  be  described  in  numbers  by  interminable 
fractions. 

In  the  ancient  system,  lines  are  only  dealt  with,  and 
every  angle  in  the  table  is  surrounded  by  lines  measur- 
ing whole  units,  and  described  by  the  use  of  a  couple  of 
simple  numbers. 

Connecting  this  with  our  present  system  of  trigo- 
nometry would  effect  a  saving  in  calculation,  and  gen- 
eral use  of  certain  peculiar  angles  by  means  of  which 
all  the  simplicity  and  beauty  of  the  work  of  the  ancients 
would  be  combined  with  the  excellences  of  our  modern 
instrumental  appliances.  Surveyors  should  appreciate 
the  advantages  to  be  derived  from  laying  out  traverses 
on  the  hypotenuses  of  "  primary"  triangles,  by  the 
saving  of  calculation  and  facility  of  plotting  to  be  ob- 
tained from  the  practice. 

The  key  to  these  old  tables  is  the  fact,  that  in  "  pri- 
mary "  triangles  the  right  angled  triangle  formed  by  the 
sine  and  versed  sine,  also  by  the  co-sine  and  co-versed- 


SOLUTION   OF   THE    PYRAMID    PROBLEM. 


sine,  is  one  in  which  base  and  perpendicular  are  meas- 
ured by  numbers  without  fractions.  These  I  will  call 
"  satellite"  triangles. 

Thus,  to  the  ''primary"  triangle  20,  21,  29,  the  ratios 
of  the  co-sinal  and  sinal  satellites  are  respectively  7  to  3, 
and  2  to  5.  (See Figure  35.)  To  the  48,  55,  73  triangle 
the  satellites  are  11,5  and  8,  3  (Fig.  62)  ;  to  the  3,  4,  5 
triangle  they  are  2,  i  and  3,  i  (Fig.  30)  ;  and  to  the  28, 
45,  53  triangle,  they  are  9,  5  and  7,  2  (Fig.  61).  The 


61. 


The  28-45:53  Triangle. 


Fig.82. 


11 


The  48-55-73  Triangle. 


primary  triangle,  7,  24,  25,  possesses  as  satellites  the 
"  primary"  triangle,  3,  4,  5,  and  the  ordinary  triangle, 
4,  i  ;  and  the  primary  triangle  41,  840,  841,  is  attended 
by  the  20,  21,  29  triangle,  as  a  satellite  with  the  ordi- 
nary triangle  41,  i,  and  so  on. 

Since  any  ratio,  however,  whose  terms,  one  or  both, 
are  represented  by  fractions,  can  be  transformed  into 
whole  numbers,  it  evidently  follows  that  every  conceiv- 


68  SOLUTION    OF   THE    PYRAMID    PROBLEM. 

able  relative  measure  of  two  lines  which  we  may  decide 
to  call  co-sine  and  co-versed-sine,  becomes  a  satellite  to 
a  corresponding  "  primary"  triangle. 

Now,  since  the  angle  of  the  satellite  on  the  circum- 
ference must  be  half  \h&  angle  of  the  adjacent  primary 
triangle  at  the  centre,  it  follows  that  in  constructing  a 
list  of  satellites  and  their  angles,  the  angles  of  the  cor- 
responding primary  triangles   can    be   found.      For  in- 
stance- 
Satellite  8,  3,  contains  20°  33'  2176" 
Satellite  2,  7,  contains  15°  56'  43-425" 

Each  of  these  angles  doubled,  gives  the  angle  of  a 
"  primary  "  triangle  as  follows,  viz.  :- 

The  48,  55,  73  triangle  ==  41°    6'  43*52" 
The  28,  45,  53  triangle  =  31°  53'  26-85" 

The  angles  of  the  satellites  together  must  always  be 
45°,  because  the  angle  at  the  circumference  of  a  quad- 
rant must  always  be  135°. 

From  the  G'izeh  plan,  as  far  as  I  have  developed  it, 
the  following  order  of  satellites  begins  to  appear,  which 
may  be  a  guide  to  the  complete  Gizeh  plan  ratio,  and 
to  those  "primary"  triangles  in  use  by  the  pyramid 
surveyors  in  their  ordinary  work. 


SOLUTION   OF   THE    PYRAMID    PROBLEM. 


69 


1 

I,        2 

2,     3 

3>     4 

4,     5 

5,     6 

6,     7 

7>     8 

3,     9 

!,        3 

2,     5 

3,     5 

4,     7 

5,     7 

7,     9 

J.     4 

2,     7 

3,     7 

4,     9 

5>     8 

J»     5 

2,     9 

3,     3 

5,     9 

7,   IJ 

i,     6 

5,   !I 

l,     7 

3,    JI 

5>   J3 

i,     8 

3'   T3 

J>     9 

i,    ii 

i,    13 

T>   '5 

Primary  triangles  may  be  found  from   the  angle  of 
the  satellite,  but    it  is  an  exceedingly  round-about  way. 
I  will,  however,  give  an  example. 

Let  us  construct  a  primary  triangle  from  the  satel- 
lite 4,  9. 


=•4444444  =  Tangt.  /  23°  57'  45-041" 


Rad.    x   4 


x  2  =  47°  5.5 '  3O'o83". 
therefore  the  angles  of  the  ''primary"  are  47°  55'  30-083". 

and  42°    4'  29*917". 

The  natural  sine  of  42°  4'  29-917"  —  '6701025. 
The  natural  co-sine  42°  4'  29-917"  —  -7422684. 


70  SOLUTION   OF    THE    PYRAMID    PROBLEM. 

The  greatest  common   measure  of    these    numbers  is 
about  102717,  therefore— 

Radius  10000000^102717  =  97 
Co-sine  7422684-^102717  =  72 
Sine  6701025  -^  102717  =  65 

and  65,  72,  97  is  the  primary  triangle  to  which  the  sat- 
ellites are  4,  9,  and  5,  13.     (See  Fig.  63.)     The  figures 


in  the  calculation  do  not  balance  exactly,  in  conse- 
quence of  the  insufficient  delicacy  of  the  tables  or  cal- 
culations. 

The  connection   between  primaries  and    satellites  is 
shown  by  figure  64. 


A  D 


Let  the  triangle  ADB  be  a  satellite,  5,  2,  which  we 
will  call  BD  20,  and  AD  8.  Let  C  be  centre  of  semi- 
circle ABE. 


SOLUTION   OF   THE    PYRAMID    PROBLEM.  71 

AD   :  DB   :  :  DB  :  DE  =50  (Euc.  VI.  8) 

AD  +  DE         =      AE  =  58  =  diameter 

AE  -4-  2  =      AC     =   BC  =  29   =  radius 

AC  ~  AD  DC  =  21    =  co-sine 

and  DB  =  20  =  sine 
From  the  preceding  it  is  manifest  that — 

sine2  ,. 
+  ver-s  =  dia. 


ver-s 


The  formula  to  find  the  "  primary  triangle"  to  any 
satellite  is  as  follows  :— 

Let  the  long  ratio  line  of  the  satellite  or  sine  be 
called  a,  and  the  short  ratio  line  or  versed-sine  be  called 
6.  Then- 

(1)  a  =   sine, 

(2)  - ± =   radius. 

2b 


(3)     ~^-^     =  c°-sine- 


Therefore    various    primary  triangles    can    be    con- 
structed on  a  side  DB  (Fig.  64)  as  sine,  by  taking  dif- 
ferent measures  for  AD  as  versed-sine. 
For  example — 


From 
Satellite 


5  =  sine      =     5 

5*     +      ^ 
2     x     i        =  radius  =    13 

£_-_£ 

"2     x     i        =  c°-s-      =    I2 


SOLUTION   OF   THE   PYRAMID    PROBLEM. 


From 
Satellite 

5,       2. 

52 

=  sine  =  5 

20 
X     1 

1     t 

^             2X2 

n                                      o 

52     -     2- 

—  radius  =7^ 

=  co-s.  =  5/4^ 

2X2 

Finally  arises  the  following  simple  rule  for  the  con- 
struction of  "  primaries"  to  contain  any  angle — Decide 
upon  a  satellite  which  shall  contain  half  the  angle — say, 
5,  i.  Call  the  first  figure  a,  the  second  b,  then— 


+  b2 
-  b2 
x  2b 


hypotenuse. 

perpendicular. 

base. 


•PRIMARY"  LOWEST  RATIO. 


5°       +       '' 

26    = 

13 

Satellite  5,   i 

5"       -       '*       = 

24    — 

12 

5     x     2X1    = 

10        = 

5 

52       +       2"       = 

29    = 

29 

Satellite  5,  2 

r2                        o2          — 

5                    2 

21        = 

21 

5X2X2    = 

2O       = 

20 

Thus— 


and — 


Having  found  the  lowest  ratio  of  the  three  sides  of 
a  "primary"  triangle,  the  lowest  whole  numbers  for 
tangent,  secant,  co-secant,  and  co-tangent,  if  required, 
are  obtained  in  the  following  manner. 

Take  for  example  the  20,  21,  29  triangle,  now 
20  x  21  =  420,  and  29  x  420  =  12180,  a  new  radius  in- 
stead of  29  from  which  with  the  sine  20,  and  co-sine  21, 
increased  in  the  same  ratio,  the  whole  canon  of  the 
20,  21,  29  triangle  will  come  out  in  whole  numbers. 


SOLUTION   OF   THE   PYRAMID   PROBLEM.  73 

Similarly  in  the  triangle  48,  55,  73,  radius  73  x  13200 
(the  product  of  48  x  55)  makes  radius  in  whole  numbers 
963600,  for  an  even  canon  without  fractions.  This  is 
because  sine  and  co-sine  are  the  two  denominators  in 
the  fractional  parts  of  the  other  lines  when  worked  out 
at  the  lowest  ratio  of  sine,  co-sine,  and  radius. 

After  I  found  that  the  plan  of  the  GTzeh  group  was 
a  system  of  "  primary  "  triangles,  I  had  to  work  out  the 
rule  for  constructing  them,  for  I  had  never  met  with  it 
in  any  book,  but  I  came  across  it  afterwards  in  the 
"  Penny  Encyclopedia,"  and  in  Rankine's  "  Civil  Engi- 
neering." 

The  practical  utility  of  these  triangles,  however,  does 
not  appear  to  have  received  sufficient  consideration.  I 
certainly  never  met  with  any  except  the  3,  4,  5,  in  the 
practice  of  any  surveyor  of  my  acquaintance. 

(For  squaring  off  a  line  nothing  could  be  more  con- 
venient than  the  20,  21,  29  triangle;  for  instance,  tak- 
ing a  base  of  40  links,  then  using  the  whole  chain  for 
the  two  remaining  sides  of  42  and  58  links.) 


SOLUTION    OF   THE    PYRAMID    PROBLEM. 


Table   of  Some    Primary   Triangles   and   their   Satellites. 


ANGLE  OF  PRIMARY 
DEC.      MIX.        SEC. 

PRIMARY 
RAD.     CO.-S.     SINE. 

! 

SATELLITE. 

ANGLE  OF  SATELLITE 
DEC.     MIX.                SEC. 

2 

47 

39'7° 

841 

840 

41 

41 

i 

j 

23 

49'85 

6 

43 

58-62 

J45 

144 

17 

17 

i 

3 

21 

59*31 

8 

47 

50-69 

85 

84 

13 

J3 

i 

4 

23 

55'34 

10 

23 

19-89 

61 

60 

II 

ii 

i 

5 

I  I 

39'94 

12 

40 

49'37 

4i 

40 

9 

9 

i 

6 

20 

24-68 

M 

14 

59'10 

65 

63 

16 

8 

i 

7 

7 

29'55 

16 

J5 

3673 

25 

24 

7 

7 

i 

8 

7 

48-36 

18 

55 

2871 

37 

35 

12 

6 

i 

9 

27 

44'35 

22 

37 

11-51 

J3 

12 

5 

5 

i 

ii 

18 

35'75 

25 

3 

27-27 

85 

77 

36 

9 

2 

12 

3i 

43'63 

25      59 

2I'22 

89 

80 

39 

13 

3 

12 

59 

40*61 

28 

4 

20*94 

«7 

'5 

8 

4 

i 

14 

2 

10-47 

30 

3o 

36H9 

65 

56 

33 

ii 

3 

15 

15 

18-24 

31      53 

26-85 

53 

45 

28 

7 

2 

15 

56 

43'42 

36 

52 

II-65 

5 

4 

3 

3 

I 

18 

26 

5-82 

4i 

6 

43*52 

73 

55 

48 

8 

3 

20 

33 

2176 

42 

4 

30-08 

97 

72 

65 

13 

5 

21 

2 

i5'°4 

43 

36 

Io'r5  j 

29 

21 

20 

5 

2 

21 

48 

5'°7 

SOLUTION  OF  THE  PYRAMID  PROBLEM. 


Table   of  some   Primary   Triangles   and   their   Satellites. 

(Continued.} 


ANGLE  OF  PRIMARY 
DEC.   MIN.   SEC. 

PRIMARY 
RAD.   CO.-S.  SINE. 

SATELLITE 

ANGLE  OF  SATELLITE 
DEG.  MIN.      SEC. 

46 

23 

49'85 

29 

20 

21 

7 

3 

23 

n 

54-92 

47 

55 

29-92 

97 

65 

72 

9 

4 

23 

57 

44-96 

48 

53 

16*48 

73 

48 

55 

ii 

5 

24 

26 

38-24 

53 

7 

48-35 

5 

3 

4 

2 

i 

26 

33 

54-I7 

58 

6 

33T5 

53 

28 

45 

9 

5 

29 

3 

l6'57 

59 

29 

23-51 

65 

33 

56 

7 

4 

29 

44 

4i'75  | 

61 

55 

39"°6 

i7 

8 

15  ! 

5 

3 

30 

57 

49'53 

64 

0 

38-78 

89 

39 

80 

8 

5 

32 

o 

19*39 

64 

56 

32-73 

85 

36 

77 

ii 

7 

32 

28 

16-36 

67 

22 

48-49  | 

13 

5 

12 

3 

2 

33 

4i 

24.24 

71 

4 

31-29 

37 

12 

35 

7 

5 

35 

32 

15-64 

73 

44 

23-27 

25 

7 

24 

4 

3 

36 

52 

11-63 

75 

45 

0*90 

65 

16 

63 

9 

7 

37 

52 

30-45 

77 
i 
79 

19 
36 

10-63  | 
40*1  1 

61 

9 
ii 

40 
60 

5 
6 

4 

5 

38 
39 

39 

48 

35-3* 
20*05 

81 

12 

9'3i 

85 

13 

84 

7 

6 

40 

36 

4-65 

83 

16 

1-38 

145 

17 

144 

9 

8 

4i 

38 

0*69 

87 

12 

20-30 

841 

4. 

840 

21 

20 

43 

36 

10*15 

76  SOLUTION   OF   THE    PYRAMID   PROBLEM. 

Reference  to  the  plan  ratio  table  at  the  commence- 
ment, and  to  the  tables  here  introduced,  will  shew  that 
most  of  the  primary  triangles  mentioned  are  indicated 
on  the  plan  ratio  table  principally  by  the  lines  corre- 
sponding to  the  ratios  of  the  satellites.  Thus— 

PRIMARY  TRIANGLE  INDICATED  BY 

17,  144,  145.  Triangle  FP,  PA,  AF  on  plan. 

13,  84>  85.  Plan  ratio  of  SJ  to  SU,  7  to  6. 

ii,  60,  61.  Plan  ratio  BC  to  FB,  6  to  5,  and  DN  to  NR,  61 

to  60. 
I2>  35>  37-  Plan  rati°  EO  to  AY,  37   to   12,  and  EA  to  AY, 

35  to  12. 
5,   12,   13.  Plan  ratio  CY  to  BC,  3  to  2  ;  JE  to  EX,  3  to  2; 

CA  to  YA,  5  to  i  ;  and  NZ  to  ZA,  12  to  5. 
8,   15,   17.  Plan  ratio  FB  to  BY,  5  to  3,  and   AC  to   BC,  15 

to  8. 
33>  56>  55-  P!an  ratio  YX  to  AY,  7  to  4  ;  AB  to  BO,  7  to  4  ; 

and  EA  to  AZ,  7  to  4. 

28,  45,  53-  Exists  on  plan,  AB,  BJ,  JA. 

3,     4?     5-  Pervades  the  plan,  and  is  also  indicated  by  plan 

ratio  GX  to   DG,  2  to  i  ;  SU  to  SV,  2  to  i  ; 

and  CY  to  YZ,  3  to  i. 

48>  55,  73-  Exists  on  plan,  FW,  WV,  VF— and  is  also  indi- 

cated by  plan  ratio  FO  to  OZ,  8  to  3. 

65,  72,  97.  Plan  ratio  AC  to  CH,  9  to  4  ;  MY  to  YZ,  9  to  4. 

20,  21,  29.  Exists  on  plan  FB,  BA,  AF  ;  and  plan  ratio,  GU 

to  DG,  5  to  2. 

It  seems  probable  that  could  I  add  to  my  pyramid 
plan  the  lines  and  triangles  that  the  missing  eleven  pyra- 
mids would  supply,  it  would  comprise  a  complete  table 
on  which  would  appear  indications  of  all  the  ratios  and 


SOLUTION   OF   THE   PYRAMID   PROBLEM.  77 

triangles  made  use  of  in   right-angled  trigonometry,  a 
"  ratiometer"  in  fact. 

I  firmly  believe  that  so  far  as  I  have  gone  it  is  cor- 
rect— and  it  is  possible,  therefore,  with  the  start  that  I 
have  made,  for  others  to  continue  the  work,  and  add  the 
eleven  pyramids  to  the  plan  in  their  correct  geometri- 
cal position.  By  continuing  the  system  of  evolution  by 
which  I  defined  the  position  of  Cephren,  and  the  little 
pyramid  to  the  south-east  of  Cheops,  after  I  had  ob- 
tained Cheops  and  Mycerinus,  may  be  rebuilt,  at  one 
and  the  same  time,  a  skeleton  of  the  trigonometrical 
tables  of  a  forgotten  civilization,  and  the  plan  of  those 
pyramids  which  are  its  only  link  with  the  present  age. 

§    13.     THE    SIZE   AND    SHAPE    OF    THE    PYRAMIDS    IN- 
DICATED   BY    THE    PLAN. 

I  pursued  my  investigations  into  the  slopes  and  alti- 
tudes of  the  pyramids  without  reference  to  the  plan, 
after  once  deciding  their  exact  bases. 

Now  it  will  be  interesting  to  note  some  of  the  ways 
in  which  the  plan  hints  at  the  shape  and  size  of  these 
pyramids,  and  corroborates  my  work. 

The  dimensions  of  Cheops  are  indicated  on  the  plan 
by  the  lines  EA  to  YA,  measuring  840  and  288  R.B. 
cubits  respectively,  being  the  half  periphery  of  its  hori- 
zontal section  at  the  level  of  Cephren's  base,  and  its 
own  altitude  from  its  own  base.  (See  Fig.  5.) 

The  line  EA,  in  fact,  represents  in  R.B.  cubits  the 
half  periphery  of  the  bases  of  either  Cheops  or  Cephren 


78  SOLUTION   OF   THE   PYRAMID   PROBLEM. 

measured  at  the  level  which  I  have  set  forth  as  the 
plan  level,  viz.,  base  of  Cephren. 

The  ratio  of  Cephren's  base  to  Cephren's  altitude  is 
indicated  on  the  plan  by  the  ratios  of  the  lines  BC  to 
EB,  or  FO  to  OR,  viz.,  32  to  21.  (See  Fig.  4.) 

The  altitude  of  Mycerinus  above  Cephren's  base 
appears  on  plan  in  the  line  EF,  measuring  136  R.B. 
cubits. 

The  line  EO  on  plan  measures  888  cubits,  which 
would  be  the  length  of  a  line  stretched  from  the  apex 
of  Cheops  to  the  point  E,  at  the  level  of  Cheops'  base. 

This  merits  consideration  : — the  lines  EA  and  AY 
are  connected  on  plan  at  the  centre  of  Cheops,  and  the 
lines  EO  and  EA  are  connected  on  plan  at  the  point  E. 

Now  the  lines  EO,  EA  and  AY  are  sides  of  a  "pri- 
mary triangle,"  whose  ratio  is  37,  35,  12,  and  whose 
measure  in  cubits  is  888,  840,  and  288  ;  and  if  we  sup- 
pose the  line  E  A  to  be  stretched  horizontally  beneath 
the  pyramids  at  the  level  of  the  base  of  Cheops  from 
E  to  A  on  plan,  and  the  line  AY  to  be  a  plumb  line 
hanging  from  the  apex  of  Cheops  to  the  level  of  his 
base,  then  will  the  line  EO  just  stretch  from  the  point 
E  to  the  apex  of  Cheops,  and  the  three  lines  will  con- 
nect the  two  main  pyramids  by  a  vertical  triangle  of 
which  EA,  AY  and  EO  form  the  base,  perpendicular, 
and  hypotenuse.  Or,  to  explain  it  in  another  manner  : 
let  the  line  EA  be  a  cord  stretching  horizontally  from 
A  at  the  centre  of  the  base  of  Cheops  to  the  point  E, 
both  ends  being  at  the  same  level ;  let  the  line  AY  be  a 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  79 

rod,  lift  it  on  the  end  A  till  it  stands  erect,  then  is  the 
end  Y  the  apex  of  Cheops,  Now,  the  line  EO  would 
just  stretch  from  the  top  of  the  rod  AY  to  the  point 
E  first  described. 

It  is  a  singular  coincidence,  and  one  that  may  be 
interesting  to  students  of  the  interior  of  the  Pyramids, 
that  the  side  EP,  of  the  small  3,  4,  5  triangle,  EP, 
PF,  FE,  in  the  centre  of  the  plan,  measures  8i'6o  R.B» 
cubits,  which  is  very  nearly  eight  times  the  ''true 
breadth "  of  the  King's  chamber  in  Cheops,  according 

to  Piazzi  Smyth;  for  — ~ — =  10*20  R.B.  cubits,  or 
206*046  pyramid  inches  (one  R.B.  cubit  being  20*2006 
pyramid  inches).  The  sides  of  this  little  triangle  meas- 
ure 8 1 -60,  1 08 '80,  and  136,  R.B.  cubits  respectively,  as 
can  be  easily  proved  from  the  plan  ratio  table. 

§  14.     A  SIMPLE  INSTRUMENT  FOR  LAYING  OFF  "  PRI- 
MARY TRIANGLES." 

A  simple  instrument  for  laying  off  "  primary  tri- 
angles" upon  the  ground,  might  have  been  made  with 
three  rods  divided  into  a  number  of  small  equal  divi- 
sions, with  holes  through  each  division,  which  rods 
could  be  pinned  together  triangularly,  the  rods  working 
as  arms  on  a  flat  table,  and  the  pins  acting  as  pointers 
or  sights. 

One  of  the  pins  would  be  permanently  fixed  in  the 
table  through  the  first  hole  of  two  of  the  rods  or  arms, 
and  the  two  other  pins  would  be  movable  so  as  to  fix 


80  SOLUTION   OF  THE    PYRAMID   PROBLEM. 

the  arms  into  the  shape  of  the  various  "  primary  tri- 
angles." 

Thus  with  the  two  main  arms  pinned  to  the  cross 
arm  in  the  2ist  and  2gth  hole  from  the  permanently 
pinned  end,  with  the  cross  arm  stretched  to  twenty  di- 
visions, a  20,  21,  29  triangle  would  be  the  result,  and 
so  on. 

§14*.  GENERAL  OBSERVATIONS. 
i 
I   must  be  excused  by  geometricians   for  going  so 

much  in  detail  into  the  simple  truths  connected  with 
right-angled  trigonometry.  M-y  object  has  been  to 
make  it  very  clear  to  that  portion  of  the  public  not 
versed  in  geometry,  that  the  Pyramids  of  Egypt  must 
have  been  used  for  land  surveying  by  right-angled  tri- 
angles with  sides  having  whole  numbers. 

A  re-examination  of  these  pyramids  on  the  ground 
with  the  ideas  suggested  by  the  preceding  pages  in 
view,  may  lead  to  interesting  discoveries. 

For  instance,  it  is  just  possible  that  the  very  accu- 
rately and  beautifully  worked  stones  in  the  walls  of  the 
King's  chamber  of  Cheops,  may  be  found  to  indicate 
the  ratios  of  the  rectangles  formed  by  the  bases  and  per- 
pendiculars of  the  triangulations  used  by  the  old  survey- 
ors— that  on  these  walls  may  be  found,  in  fact,  corrob- 
oration  of  the  theory  that  I  have  set  forth.  I  am  led 
to  believe  also  from  the  fact  that  G'izeh  was  a  central 
and  commanding  locality,  and  that  it  was  the  custom  of 
those  who  preceded  those  Egyptians  that  history  tells 


SOLUTION  OF  THE  PYRAMID  PROBLEM.         8 1 

of,  to  excavate  mighty  caverns  in  the  earth — that,  there- 
fore, in  the  limestone  upon  which  the  pyramids  are 
built,  and  underneath  the  pyramids,  may  be  found  vast 
excavations,  chambers  and  galleries,  that  had  entrance 
on  the  face  of  the  ridge  at  the  level  of  High  Nile. 
From  this  subterraneous  city,  occupied  by  the  priests 
and  the  surveyors  of  Memphis,  access  may  be  found  to 
every  pyramid ;  and  while  to  the  outside  world  the 
pyramids  might  have  appeared  sealed  up  as  mauso- 
leums to  the  Kings  that  it  may  have  seen  publicly  in- 
terred therein,  this  very  sealing  and  closing  of  the  outer 
galleries  may  have  only  rendered  their  mysterious 
recesses  more  private  to  the  priests  who  entered  from 
below,  and  who  were,  perhaps,  enabled  to  ascend  by  pri- 
vate passages  to  their  very  summits.  The  recent  dis- 
covery of  a  number  of  regal  mummies  stowed  away  in 
an  out  of  the  way  cave  on  the  banks  of  the  Nile,  points 
to  the  unceremonious  manner  in  which  the  real  rulers 
of  Kings  and  people  may  have  dealt  with  their  sov- 
ereigns, the  pomp  and  circumstance  of  a  public  burial 
once  over.  It  is  just  possible  that  the  chambers  in  the 
pyramids  may  have  been  used  in  connection  with  their 
mysteries:  and  the  small  passages  called  by  some  "ven- 
tilators" or  "air  passages,"  sealed  as  they  were  from 
the  chamber  by  a  thin  stone  (and  therefore  no  ventila- 
tors) may  have  been  auditory  passages  along  which 
sound  might  have  been  projected  from  other  chambers 
not  yet  opened  by  the  moderns ;  sounds  which  were 
perhaps  a  part  of  the  "  hanky  panky"  of  the  ancient 


82         SOLUTION  OF  THE  PYRAMID  PROBLEM. 

ceremonial  connected  with  the  "mysteries"  or  the  "  re- 
ligion "  of  that  period. 

Down  that  "well "  which  exists  in  the  interior  of 
Cheops,  and  in  the  limestone  foundations  of  the  pyra- 
mid, should  I  be  disposed  to  look  for  openings  into  the 
vast  subterraneous  chambers  which  I  am  convinced  do 
exist  below  the  Pyramids  of  G'fzeh. 

The  priests  of  the  Pyramids  of  Lake  Moeris  had 
their  vast  subterranean  residences.  It  appears  to  me 
more  than  probable  that  those  of  G'fzeh  were  similarly 
provided.  And  I  go  further : — Out  of  these  very 
caverns  may  have  been  excavated  the  limestone  of 
which  the  pyramids  were  built,  thus  killing  two  birds 
with  one  stone — building  the  instruments  and  finding 
cool  quarters  below  for  those  who  were  to  make  use  of 
them.  In  the  bowels  of  that  limestone  ridge  on  which 
the  pyramids  are  built  will  yet  be  found,  I  feel  con- 
vinced, ample  information  as  to  their  uses.  A  good 
diamond  drill  with  two  or  three  hundred  feet  of  rods  is 
what  is  what  is  wanted  to  test  this,  and  the  solidity  of 
the  pyramids  at  the  same  time. 

§  15.     PRIMARY    TRIANGULATION. 

Primary  triangulation  would  be  useful  to  men  of 
almost  every  trade  and  profession  in  which  tools  or 
instruments  are  used.  Any  one  might  in  a  short  time 
construct  a  table  for  himself  answering  to  every  degree 
or  so  in  the  circumference  of  a  circle  for  which  only 
forty  or  fifty  triangles  are  required. 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  83 

It  would  be  worth  while  for  some  one  to  print  and 
publish  a  correct  set  of  these  tables  embracing  a  close 
division  of  the  circle,  in  which  set  there  should  be  a 
column  showing  the  angle  in  degrees,  minutes,  seconds 
and  decimals,  and  also  a  column  for  the  satellite,  thus — 


SATELLITE. 


5     2  20     21      29          43°  36'   10-15" 

73  21     20     29          46°  23'  49-85" 

and  so  on.  Such  a  set  of  tables  would  be  a  boon  to 
sailors,  architects,  surveyors,  engineers,  and  all  handi- 
craftsmen :  and  I  make  bold  to  say,  would  assist  in  the 
intricate  investigations  of  the  astronomer:  —  and  the 
rule  for  building  the  tables  is  so  simple,  that  they  could 
easily  be  achieved.  The  architect  from  these  tables 
might  arrange  the  shape  of  his  chambers,  passages  or 
galleries,  so  that  all  measures,  not  only  at  right  angles 
on  the  walls,  but  from  any  corner  of  floor  to  ceiling 
should  be  even  feet.  The  pitch  of  his  roofs  might  be 
more  varied,  and  the  monotony  of  the  buildings  relieved, 
with  rafters  and  tie-beams  always  in  even  measures. 
The  one  solitary  3,  4,  5  of  Vitruvius  would  cease  to  be 
his  standard  for  a  staircase  ;  and  even  in  doors  and 
sashes,  and  panels  of  glass,  would  he  be  alive  to  the 
perfection  of  rectitude  gained  by  evenly-measured  di- 
agonals. By  a  slight  modification  of  the  compass  card, 
the  navigator  of  blue  water  might  steer  his  courses  on 
the  hypotenuses  of  great  primary  triangles — such  tables 
would  be  useful  to  all  sailors  and  surveyors  who  have 
to  deal  with  latitude  and  departure.  For  instance,  fa- 


84  SOLUTION   OF  THE  PYRAMID    PROBLEM. 

miliarity  with  such  tables  would  make  ever  present  in 
the  mind  of  the  surveyor  or  sailor  his  proportionate 
northing  and  easting,  no  matter  what  course  he  was 
steering  between  north  and  east,  "the  primary"  em- 
braces the  three  ideas  in  one  view. 

In  designing  trussed  roofs  or  bridges,  the  "  pri- 
maries" would  be  invaluable  to  the  engineer,  strain-cal- 
culations on  diagonal  and  upright  members  would  be 
simplified,  and  the  builder  would  find  the  benefit  of  a 
measure  in  even  feet  or  inches  from  centre  of  one  pin 
or  connection  to  another. 

For  earthwork  slopes  3,  4,  5 ;  20,  21,  29;  21,  20,  29  ; 
and  4,  3,  5  would  be  found  more  convenient  ratios  than 
i  /<?  i,  and  i^  to  i,  etc.  Templates  and  battering  rules 
would  be  more  perfect  and  correct,  and  the  engineer 
could  prove  his  slopes  and  measure  his  work  at  one 
and  the  same  time  without  the  aid  of  a  staff  or  level ; 
the  slope  measures  would  reveal  the  depth,  and  the 
slope  measures  and  bottom  width  would  be  all  the 
measures  required,  while  the  top  width  would  prove  the 
correctness  of  the  slopes  and  the  measurements. 

To  the  land  surveyor,  however,  the  primary  triangle 
would  be  the  most  useful,  and  more  especially  to  those 
laying  out  new  holdings,  whether  small  or  large,  in  new 
countries. 

Whether  it  be  for  a  "squatter's  run,"  or  for  a  town 
allotment,  the  advantages  of  a  diagonal  measure  to 
every  parallelogram  in  even  miles,  chains,  or  feet,  should 
be  keenly  felt  and  appreciated. 


SOLUTION   OF  THE    PYRAMID    PROBLEM.  85 

This  was,  I  believe,  one  of  the  secrets  of  the  speedy 
and  correct  replacement  of  boundary  marks  by  the 
Egyptian  land  surveyors. 

I  have  heard  of  a  review  in  the  "  Contemporary," 
September,  1881,  referring  to  the  translation  of  a  papy_ 
rus  in  the  British  Museum,  by  Dr.  Eisenlohr — "A 
handbook  of  practical  arithmetic  and  geometry  "etc.,  "such 
as  we  might  suppose  would  be  used  by  a  scribe  acting  as 
clerk  of  the  works,  or  by  an  architect  to  shew  the  work- 
ing out  of  the  problems  he  had  to  solve  in  his  operations" 
I  should  like  to  see  a  translation  of  the  book,  from  which 
it  appears  that  "  the  chtmsiness  of  the  Egyptian  method 
is  very  remarkable"  Perhaps  this  Egyptian  "Hand- 
book" may  yet  shew  that  their  operations  were  not  so 
"  clumsy"  as  they  appear  at  first  sight  to  those  accus- 
tomed to  the  practice  of  modern  trigonometry.  I  may 
not  have  got  the  exact  "  hang"  of  the  Egyptian  method 
of  land  surveying — for  I  do  not  suppose  that  even  their 
"  clumsy  "  method  is  to  be  got  at  intuitively  ;  but  I  claim 
that  I  have  shewn  how  the  Pyramids  could  be  used  for 
that  purpose,  and  that  the  subsidiary  instrument  de- 
scribed by  me  was  practicable. 

I  claim,  therefore,  that  the  theory  I  have  set  up, 
that  the  pyramids  were  the  theodolites  of  the  Egyp- 
tians, is  sound.  That  the  ground  plan  of  these  pyra- 
mids discloses  a  beautiful  system  of  primary  triangles 
and  satellites  I  think  I  have  shown  beyond  the  shadow 
of  a  doubt ;  and  that  this  system  of  geometric  triangu- 
lation  or  right-angled  trigonometry  was  the  method  prac- 


86  SOLUTION    OF    THE    PYRAMID    PROBLEM. 

tised,  seems  in  the  preceding  pages  to  be  fairly  estab- 
lished. I  claim,  therefore,  that  I  have  discovered  and 
described  the  main  secret  of  the  pyramids,  that  I  have 
found  for  them  at  last  a  practical  use,  and  that  it  is  no 
longer  "a  marvel  how  after  the  annual  inundation,  each 
property  could  have  been  awirately  described  by  the  aid 
of  geometry  "^-\  have  advanced  nothing  in  the  shape  of 
a  theory  that  will  not  stand  a  practical  test ;  but  to  do 
it,  the  pyramids  should  be  re-cased.  Iron  sheeting,  on 
iron  or  wooden  framework,  would  answer.  I  may  be 
wrong  in  some  of  my  conclusions,  but  in  the  main  I  am 
satisfied  that  I  am  right.  It  must  be  admitted  that  I 
have  worked  under  difficulties  ;  a  glimpse  at  the  pyra- 
mids three  and  twenty  years  ago,  and  the  meagre  library 
of  a  nomad  in  the  Australian  wilderness  having  been  all 
my  advantages,  and  time  at  my  disposal  only  that 
snatched  from  the  rare  intervals  of  leisure  afforded  by 
an  arduous  professional  life. 

After  fruitless  waiting  for  a  chance  of  visiting  Egypt 
and  Europe,  to  sift  the  matter  to  the  bottom,  I  have  at 
last  resolved  to  give  my  ideas  to  the  world  as  they  stand  ; 
crude  necessarily,  so  I  must  be  excused  if  in  some  de- 
tails I  may  be  found  erroneous  ;  there  is  truth  I  know 
in  the  general  conclusions.  I  am  presumptuous  enough 
to  believe  that  the  R.B.  cubit  of  1:685  British  feet  was 
the  measure  of  the  pyramids  of  G'fzeh,  although  there 
may  have  been  an  astronomical  25  inch  cubit  also.  It 
appears  to  me  that  no  cubit  measure  to  be  depended  on 
is  either  to  be  got  from  a  stray  measuring  stick  found 


SOLUTION    OF   THE  PYRAMID    PROBLEM.  S/ 

in  the  joints  of  a  ruined  building,  or  from  any  line  or 
dimensions  of  one  of  the  pyramids.  I  submit  that  a 
most  reasonable  way  to  get  a  cubit  measure  out  of  the 
Pyramids  of  G'fzeh,  was  to  do  as  I  did  : — take  them  as  a 
whole,  comprehend  and  establish  the  general  ground 
plan,  find  it  geometric  and  harmonic,  obtain  the  ratios 
of  all  the  lines,  establish  a  complete  set  of  natural  and 
even  numbers  to  represent  the  measures  of  the  lines, 
and  finally  bring  these  numbers  to  cubits  by  a  common 
multiplier  (which  in  this  case  was  the  number  eight). 
After  the  whole  proportions  had  been  thus  expressed 
in  a  cubit  evolved  from  the  whole  proportions,  I  estab- 
lished its  length  in  British  feet  by  dividing  the  base  of 
Cephren,  as  known,  by  the  number  of  my  cubits  repre- 
senting its  base.  It  is  pretty  sound  evidence  of  the 
theory  being  correct  that  this  test,  with  420  cubits  neat 
for  Cephren,  gave  me  also  a  neat  measure  for  Cheops, 
from  Piazzi  Smyth's  base,  of  452  cubits,  and  that  at  the 
same  level,  these  two  pyramids  become  equal  based. 

I  have  paid  little  attention  to  the  inside  measure- 
ments. I  take  it  we  should  first  obtain  our  exoteric 
knowledge  before  venturing  on  esotoric  research.  Thus 
the  intricate  internal  measurements  of  Cheops,  made  by 
various  enquirers  have  been  little  service  to  me,  while 
the  accurate  measures  of  the  base  of  Cheops  by  Piazzi 
Smyth,  and  John  James  Wild's  letter  to  Lord  Brough- 
am, helped  me  amazingly,  as  from  the  two  I  estab- 
lished the  plan  level  and  even  bases  of  Cheops  and 
Cephren  at  plan  level — as  I  have  shown  in  the  preced- 


88  SOLUTION    OF   THE    PYRAMID    PROBLEM. 

ing  pages.  My  theory  demanded  that  both  for  the 
building  of  the  pyramids  and  for  the  construction  of  the 
models  or  subsidiary  instruments  of  the  surveyors,  sim- 
ple slope  ratios  should  govern  each  building ;  before  I 
conclude,  I  shall  show  how  I  got  at  my  slope  ratios,  by 
evolving  them  from  the  general  ground  plan. 

I  am  firmly  convinced  that  a  careful  investigation 
into  the  ground  plans  of  the  various  other  groups  of 
pyramids  will  amply  confirm  my  survey  theory — the  rel- 
ative positions  of  the  groups  should  also  be  established 
—much  additional  light  will  be  then  thrown  on  the 
subject. 

Let  me  conjure  the  investigator  to  view  these  piles 
from  a  distance  with  his  mind's  eye,  as  the  old  survey- 
ors viewed  them  with  their  bodily  eye.  Approach  them 
too  nearly,  and,  like  Henry  Kinglake,  you  will  be  lost 
in  the  "one  idea  of  solid  immensity."  Common  sense 
tells  us  they  were  built  to  be  viewed  from  a  distance. 

Modern  surveyors  stand  near  their  instruments,  and 
send  their  flagmen  to  a  distance  ;  the  Egyptian  sur- 
veyor was  one  of  his  own  flagmen,  and  his  instruments 
were  towering  to  the  skies  on  the  distant  horizon. 
These  mighty  tools  will  last  out  many  a  generation  of 
surveyors. 

The  modern  astronomer  from  the  top  of  an  observa- 
tory points  his  instruments  direct  at  the  stars ;  the 
Egyptian  astronomer  from  the  summit  of  his  particular 
pyramid  directed  his  observations  to  the  rising  and  set- 
ting of  the  stars,  or  the  positions  of  the  heavenly  bodies 


SOLUTION    OF   THE    PYRAMID    PROBLEM.  89 

in  respect  to  the  far  away  groups  of  pyramids  scattered 
around  him  in  the  distance  ;  and  by  comparing  notes, 
and  with  the  knowledge  of  the  relative  position  of  the 
groups,  did  these  observers  map  out  the  sky.  Solar  and 
lunar  shadows  of  their  own  pyramids  on  the  flat 
trenches  prepared  for  the  purpose,  enabled  the  astrono- 
mer at  each  observatory  to  record  the  annual  and 
monthly  flight  of  time,  while  its  hours  were  marked  by 
the  shadows  of  their  obelisks,  capped  by  copper 
pyramids  or  balls,  on  the  more  delicate  pavements 
of  the  court-yards  of  their  public  buildings. 

We  must  grasp  that  their  celestial  and  terrestrial 
surveys  were  almost  a  reverse  process  to  our  own, 
before  we  can  venture  to  enquire  into  its  details.  It 
then  becomes  a  much  easier  tangle  to  unravel.  That  a 
particular  pyramid  among  so  many,  should  have  been 
chosen  as  a  favoured  interpreter  of  Divine  truths,  seems 
an  unfair  conclusion  to  the  other  pyramids  ; — that  the 
other  pyramids  were  rough  and  imperfect  imitations, 
appears  to  my  poor  capacity  "  a  base  and  impotent 
conclusion  ;  "•  —(as  far  as  I  can  learn,  Myceftnus,  in  its 
perfection,  was  a  marvel  of  the  mason's  art  ;)  but  that 
one  particular  pyramid  should  have  anything  to  do  with 
the  past  or  the  future  of  the  lost  ten  tribes  of  Israel 
(whoever  that  fraction  of  our  present  earthly  commu- 
nity may  be),  seems  to  me  the  wildest  conclusion  of  all, 
except  perhaps  the  theory  that  this  one  pyramid  points 
to  the  future  of  the  British  race.  Yet  in  one  way 
do  I  admit  that  the  pyramids  point  to  our  future. 


90  SOLUTION    OF    THE    PYRAMID    PROBLEM. 

Thirty-six  centuries  ago,  they,  already  venerable 
with  antiquity,  looked  proudly  down  on  living  labouring 
Israel,  in  helpless  slavery,  in  the  midst  of  an  advanced 
civilization,  of  which  the  history,  language,  and  religion 
are  now  forgotten,  or  only  at  best,  slightly  understood. 

Thirty-six  centuries  hence,  they  may  look  down  on  a 
civilization  equally  strange,  in  which  our  history,  lan- 
guage, and  religion,  Hebrew  race,  and  British  race,  may 
have  no  place,  no  part. 

If  the  thoughts  of  noble  poets  live,  as  they  seem  to 
do,  old  Cheops,  that  mountain  of  massive  masonry, 
may  (like  the  brook  of  our  Laureate),  in  that  dim 
future,  still  be  singing,  as  he  seems  to  sing  now,  this 
idea,  though  not  perhaps  these  words  : 

"  For  men  may  come,  and  men  may  go, 
But  I  go  on  for  ever." 

"  Ars  longa,  vita  brevis."  Man's  work  remains,  when 
the  workman  is  forgotten  ;  fair  work  and  square,  can 
never  perish  entirely  from  men's  minds,  so  long  as  the 
world  stands.  These  pyramids  were  grand  and  noble 
works,  and  they  will  not  perish  till  their  reputation  has 
been  re-established  in  the  world,  when  they  will  live 
in  men's  memories  to  all  generations  as  symbols  of  the 
mighty  past.  To  the  minds  of  many  now,  as  to  Jose- 
phus  in  his  day,  they  are  "  vast  and  vain  monuments? 
records  of  folly.  To  me  they  are  as  monuments  of 
peace,  civilization  and  order — relics  of  a  people  living 
under  wise  and  beneficent  rulers — evidences  of  cul- 
tivation, science,  and  art. 


SOLUTION   OF    THE    PYRAMID    PROBLEM.  91 


§   16.    THE   PENTANGLE  OR  FIVE  POINTED   STAR  THE 
GEOMETRIC    SYMBOL    OF    THE    GREAT    PYRAMID. 

From  time  immemorial  this  symbol  has  been  a 
blazing  pointer  to  grand  and  noble  truths,  and  a  solemn 
emblem  of  important  duties. 

Its  geometric  significance,  however,  has  long  been 
lost  sight  of. 

It  is  said  to  have  constituted  the  seal  or  signet 
of  King  Solomon  (1000  B.C.),  and  in  early  times  it  was 
in  use  among  the  Jews,  as  a  symbol  of  safety. 

It  was  the  Pentalpha  of  Pythagoras,  and  the  Pytha- 
gorean emblem  of  health  (530  B.C.). 

It  was  carried  as  the  banner  of  Antiochus,  King  of 
Syria  (surnamed  Soter,  or  the  Preserver),  in  his  wars 
against  the  Gauls  (260  B.C.).  Among  the  Cabalists,  the 
star  with  the  sacred  name  written  on  each  of  its  points, 
and  in  the  centre,  was  considered  talismanic  ;  and 
in  ancient  times  it  was  employed  all  over  Asia  as  a 
charm  against  witchcraft.  Even  now,  European  troops 
at  war  with  Arab  tribes,  sometimes  find,  under  the 
clothing,  on  the  breasts  of  their  slain  enemies,  this 
ancient  emblem,  in  the  form  of  a  metal  talisman,  or 
charm. 

The  European  Goethe  puts  these  words  into  the 
mouth  of  Mephistopheles  : 

"  I  am  hindered  egress  by  a  quaint  device  upon  the  threshold, 
— that  five-toed  damned  spell." 


92  SOLUTION    OF   THE    PYRAMID    PROBLEM. 

I  shall  set  forth  the  geometric  significance  of  this  star, 
as  far  as  my  general  subject  warrants  me,  and  show  that 
it  is  the  geometric  emblem  of  extreme  and  mean  ratio,  and 
the  symbol  of  the  Egyptian  Pyramid  Cheops. 

A  plane  geometric  star,  or  a  solid  geometric  pyra- 
mid, may  be  likened  to  the  corolla  of  a  flower,  each 
separate  side  representing  a  petal.  With  its  petals 
opened  and  exposed  to  view,  the  flower  appears  in  all 
its  glorious  beauty  ;  but  when  closed,  many  of  its 
beauties  are  hidden.  The  botanist  seeks  to  view  it  flat 
or  open  in  its  geometric  symmetry,  and  also  closed,  as  a 
bud,  or  in  repose  : — yet  judges  and  appreciates  the  one 
state  from  the  other.  In  the  same  manner  must  we  deal 
with  the  five  pointed  star,  and  also  with  the  Pyramid 
Cheops. 

In  dealing  with  so  quaint  a  subject,  I  may  be 
excused,  in  passing,  for  the  quaint  conceit  of  likening 
the  interior  galleries  and  chambers  of  this  pyramid 
to  the  interior  whorl  of  a  flower,  stamens  and  pistil, 
mysterious  and  incomprehensible. 

Figure  67  (page  roi),  is  the  five  pointed  star, 
formed  by  the  unlapping  of  the  five  slant  sides  of  a 
pyramid  with  a  pentagonal  base. 

Figure  70  (page  106),  is  a  star  formed  by  the  unlap- 
ping of  the  four  slant  sides  of  the  pyramid  Cheops. 

The  pentagon  GFRHO,  (Fig.  67)  is  the  base  of  the 
pyramid  "  Pentalpha"  and  the  triangles  EGF,  BFR, 
ROH,  HNQ  and  QAG,  represent  the  five  sides,  so  that 
supposing  the  lines  GF,  FR,  RH,  HQ  and  QG,  to  be 


SOLUTION    OF    THE    PYRAMID    PROBLEM.  93 

hinges  connecting  these  sides  with  the  base,  then  by 
lifting  the  sides,  and  closing  them  in,  the  points  A,  E, 
B,  O,  and  N,  would  meet  over  the  centre  C. 

Thus  do  we  close  the  geometric  flower  Pentalpha, 
and  convert  it  into  a  pyramid. 

In  the  same  manner  must  we  lift  the  four  slant  sides 
of  the  pyramid  Cheops  from  its  star  development, 
(Fig.  70)  and  close  them  in,  the  four  points  meeting 
over  the  centre  of  the  base,  forming  the  solid  pyramid. 
Such  transitions  point  to  the  indissoluble  connection 
between  plane  and  solid  geometry. 

As  the  geometric  emblem  of  extreme  and  mean  ratio, 
the  pentangle  appears  as  an  assemblage  of  lines  divided 
the  one  by  the  others  in  extreme  and  mean  ratio. 

To  explain  to  readers  not  versed  in  geometry,  what 
extreme  and  mean  ratio  signifies,  I  refer  to  Figure  65  :— 


Fig.  65. 


Let  AB  be  the  given  line  to  be  divided  in  extreme 
and  mean  ratio,  i.e.,  so  that  the  whole  line  may 
be  to  the  greater  part,  as  the  greater  is  to  the 
less  part. 

Draw  BC  perpendicular  to  AB,  and  equal  to  half  AB. 
Join  AC  ;  and  \vith  BC  as  a  radius  from  C  as  a  centre, 
describe  the  arc  DB;  then  with  centre  A,  and  radius 


94  SOLUTION    OF    THE   PYRAMID    PROBLEM. 

AD,  describe  the  arc  DE  ;  so  shall  AB  be  divided  in  E, 
in  extreme  and  mean  ratio,  or  so  that  AB  :  AE  :  :  AE  : 
EB.  (Note  that  AE  is  equal  to  the  side  of  a  deca- 
gon inscribed  in  a  circle  with  radius  AB.) 

Let  it  be  noted  that  since  the  division  of  a  line  in 
mean  and  extreme  ratio  is  effected  by  means  of  the  2,  i 
triangle,  ABC,  therefore,  as  the  exponent  of  this  ratio, 
another  reason  presents  itself  why  it  should  be  so  im- 
portant a  feature  in  the  G'fzeh  pyramids  in  addition  to 
its  connection  with  the  primary  triangle  3,  4,  5. 


Fig.  66. 


B 


To  complete  the  explanation  offered  with  figure  65, 
I  must  refer  to  Fig.  66,  where  in  constructing  a  pentagon, 
the  2,  i  triangle  ABC,  is  again  made  use  of. 

The  line  AB  is  a  side  of  the  pentagon.  The  line 
BC  is  a  perpendicular  to  it,  and  half  its  length. 
The  line  AC  is  produced  to  F,  CF  being  made 
equal  to  CB  ;  then  with  B  as  a  centre,  and  radius 
BF,  the  arc  at  E  is  described  ;  and  with  A  as  a 
centre,  and  the  same  radius,  the  arc  at  E  is  inter- 
sected, their  intersection  being  the  centre  of  the 


SOLUTION   OF    THE    PYRAMID    PROBLEM. 


95 


circle    circumscribing    the    pentagon,   and    upon 
which  the  remaining  sides  are  laid  off. 
We  will  now  refer  to  figure  67,   in  which   the   pent- 
angle  appears  as  the  symbolic  exponent  of  the  division 
of  lines  in  extreme  and  mean  ratio. 


Thus  : 


MC: 
AF: 
AB: 


MH  :  : 
AG:  : 
AF:: 


MH:  HC 
AG:  GF 
AF:  FB 


i — being  the  geo- 


while  MN,  MH  or  XC:    CD  :  :    2 
metric  template  of  the  work. 

Thus  every  line  in  this  beautiful  symbol  by  its  inter- 
sections with  the  other  lines,  manifests  the  problem. 

Note  also  that 

GH  =  GA 

AE  =  AF 
DH  =  DE 

I  append  a  table  showing  the  comparative  measures 
of  the  lines  in  Fig.  67,  taking  radius  of  the  circle  as  a 
million  units. 

.  67. 


96  SOLUTION    OF  THE    PYRAMID   PROBLEM. 

Table  Showing  the  Comparative  Measures  of  Lines. 
(Fig.  67.) 


ME  —  2000000  =  diameter. 

AB   —  1902113  =  AD  -f  DB 

MB  =  1618034  =  MC  +  MH  =  MP  +  PB 

AS    =  1538841-5 

EP    ==  1453086  =  AG  +  FB 

AF   =  1175570  —  AE  =  GB 

MC  =  i  oooooo  =  radius  =  CD  +  DX  =  CH  +  CX 

AD   =  951056-5  =  DB  =  DS 

PB    =  854102 

QS   =  812298-5 

MP  =  763932  —  CH  x  2  =  base  of  Cheops. 

AG  =  726543  =  GH  =  XH  =±  HN  =  PF  =  FB  =  Slant 

edge  of  Cheops.   =  slant  edge  of  Pent.  Pyr. 
DE   =  690983  =  DH  =  XD  =  apothem  of  Pentagonal  Pyramid. 

f  apothem  of  Cheops. 
MH  -—  618034  —  MN  =  XC  =  -(  altitude  of  Pentagonal  Pyramid. 

[  side  of  decagon  inscr'd  in  circle. 
MS  —  500000 

mean  proportional  between  MH  and  HC 


481:868  = 

(  altitude  of  Cheops. 

OP   =  449027  =  GF  =  GD  +  DF 
HC  =  381966  =  half  base  of  Cheops. 

so  =  363271-5  =  HS 

CD  =  309017  =half  M  H 
PR  =  277516 
GD  =  224513-5 
SP  =  263932 


The  triangle  DXH  represents  a  vertical  section  of 
the  pentagonal  pyramid  ;  the  edge  HX  is  equal  to  HN, 
and  the  apothem  DX  is  equal  to  DE.  Let  DH  be  a 


SOLUTION    OF    THE    PYRAMID   PROBLEM.  97 

hinge  attaching  the  plane  DXH  to  the  base,  now  lift 
the  plane  DXH  until  the  point  X  is  vertical  above  the 
centre  C.  Then  the  points  A,  E,  B,  O,  N  of  the  five 
slant  slides,  when  closed  up,  will  all  meet  at  the  point  X 
over  the  centre  C. 

We  have  now  built  a  pyramid  out  of  the  pentangle, 
whose  slope  is  2  to  i,  altitude  CX  being  to  CD  as  2 
to  i.  ' 

Apothem  DX  ==  DE 
Altitude     CX  ==  HM  or  MN 
Altitude     CX  +  CH  ==  CM  radius. 
Apothem  DX  +  CD  ==  CM  radius. 
Edge         HX  ==  HN  or  PF 

Note  also  that 


OP  ==  HR 

Let  us  now  consider  the  Pentangle  as  the  symbol  of 
the  Great  Pyramid  Cheops. 

The  line  MP  =  the  base  of  Cheops. 
The  line  CH  —  half  base  of  Cheops. 
The  line  HM  =  apothem  of  Cheops. 
The  line  HN  =  slant  edge  of  Cheops. 
Thus  :  Apothem  of  Cheops  =  side  of  decagon. 

Apothem  of  Cheops  =  altitude     of     pentagonal 
pyramid. 


98  SOLUTION   OF  THE    PYRAMID    PROBLEM. 

Slant  edge  of  Cheops  =  slant  edge  of  pentago- 
nal pyramid. 

Now  sii\ce  apothem  of  Cheops  =  MH 
and  half  base  of  Cheops  =  HC 

then  do  apothem  and  half  base  represent,  when  taken 
together,  extreme  and  mean  ratio,  and  altitude  is  a 
mean  proportional  between  them :  it  having  already 
been  stated,  which  also  is  proved  by  the  figures  in  the 
table,  that  MC  :  MH  :  :  MH  :  HC  and  apoth  :  alt  :: 
alt  :  half  base. 

Thus  is  the  four  pointed  star  Cheops  evolved  from 
the  five  pointed  star  Pentalpha.  This  is  shown  clearly 
by  Fig.  68,  thus  :— 


Within  a  circle  describe  a  pentangle,  around  the 
interior  pentagon  of  the  star  describe  a  circle,  around 
the  circle  describe  a  square  ;  then  will  the  square  repre- 
sent the  base  of  Cheops. 

Draw  two  diameters  of  the  outer  circle  passing 
through  the  centre  square  at  right  angles  to  each  other, 
and  each  diameter  parallel  to  sides  of  the  square  ;  then 


SOLUTION    OF    THE    PYRAMID    PROBLEM.  99 

will  the  parts  of  these  diameters  between  the  square 
and  the  outer  circle  represent  the  four  apothems  of  the 
four  slant  sides  of  the  pyramid.  Connect  the  angles  of 
the  square  with  the  circumference  of  the  outer  circle  by 
lines  at  the  four  points  indicated  by  the  diameters,  and 
the  star  of  the  pyramid  is  formed,  which,  when  closed 
as  a  solid,  will  be  a  correct  model  of  Cheops. 

Calling  apothem  of  Cheops,  MH  =  34 

and  half  base,  HC  =  21 

as  per  Figure  6.     Then — MH  +  MC  =  55 

and  55  :  34  :  :  34  :  21*018,  being  only  in  error  a  few 
inches  in  the  pyramid  itself,  if  carried  into  actual  meas- 
ures. 

The  ratio,  therefore,  of  apothem  to  half-base,  34  to 
21,  which  I  ascribe  to  Cheops,  is  as  near  as  stone  and 
mortar  can  be  got  to  illustrate  the  above  proportions. 

Correctly  stated  arithmetically  let  MH  =  2. 

Then  HC  =  =  YS  -  i 

MC    =:    V$     +     I 

and  altitude  of  Cheops  =  VMH  X  MC 


100 


SOLUTION   OF   THE   PYRAMID   PROBLEM. 


Let  us   now  compare  the  construction  of  the  two 
stars  :— 

E 


Fig.  69 


TO   CONSTRUCT   THE     STAR    PENTALPHA 
FIG.    69. 


Describe  a  circle. 

Draw  diameter  MCE. 

Divide  MC  in  mean  and  extreme 

ratio  at  H. 

Lay  off  half  MH  from  C,  to  D. 
Draw  chord  ADB,  at  right  angles 

to  diameter  ECM. 
Draw  chord  BHN,  through  H. 
Draw  chord  AHO,  through  H. 
Connect  NE. 
Connect  EO. 


TO    CONSTRUCT    THE    STAR    CHEOPS, 
FIG.    70. 


Describe  a  circle. 

Draw  diameter  MCE. 

Divide  MC  in  mean  and  extreme 
ratio,  at  H. 

Describe  an  inner  circle  with  ra- 
dius CH,  and  around  it  describe 
the  square  a,  b,  c,  d. 

Draw  diameter  ACB,  at  right  an- 
gles to  diameter  ECM. 

Draw  Aa,  aE,  Eb,  bB,  Bd,  dM, 
Me,  and  cA. 


The  question  now  arises,  does  this  pyramid  Cheops 
set  forth  by  the  relations  of  its  altitude  to  perimeter  of 


SOLUTION   OF  THE   PYRAMID    PROBLEM.  IQI 

base  the  ratio  of  diameter  to  circumference  ;  or,  does  it 
set  forth  mean  proportional,  and  extreme  and  mean 
ratio,  by  the  proportions  of  its  apothem,  altitude,  and 
half-base  ?  The  answer  is — from  the  practical  impossi- 
bility of  such  extreme  accuracy  in  such  a  mass  of 
masonry,  that  it  points  alike  to  all,  and  may  as  fairly  be 
considered  the  exponent  of  the  one  as  of  the  others. 
Piazzi  Smyth  makes  Cheops  76 1-65  feet  base,  and 
484*91  feet  altitude,  which  is  very  nearly  what  he  calls 
a  n  pyramid,  for  which  I  reckon  the  altitude  would  be 
about  484*87  feet  with  the  same  base  :  and  for  a  pyr- 
amid of  extreme  and  mean  ratio  the  altitude  would  be 
484-34  feet. 

The  whole  difference,  therefore,  is  only  about  six 
inches  in  a  height  of  nearly  five  hundred  feet.  This 
difference,  evidently  beyond  the  power  of  man  to  dis- 
cover, now  that  the  pyramid  is  a  ruin,  would  even  in  its 
perfect  state  have  been  inappreciable. 

It  appears  most  probable  that  the  star  Pentalpha 
led  to  the  star  Cheops,  and  that  the  star  Cheops 
{Fig.  70)  was  the  plan  used  by  the  ancient  architect, 
and  the  ratio  of  34  to  21,  hypotenuse  to  base,  the  tem- 
plate used  by  the  ancient  builders. 

Suppose  some  king  said  to  his  architect,  "  Make  me 
a  plan  of  a  pyramid,  of  which  the  base  shall  be  420 
cubits  square,  and  altitude  shall  be  to  the  perimeter  of 
the  base  as  the  radius  of  a  circle  to  the  circumference." 
-Then  might  the  architect  prepare  an  elaborate  plan 
in  which  the  relative  dimensions  would  be  about — 


102  SOLUTION   OF  THE    PYRAMID    PROBLEM. 


R.    B.    CUBITS. 


f  Base 420 

Base  angle  51°  51'  i4*3''-<  Altitude.  .  . .  267*380304  &c. 

[  Apothem 339'988573  &c- 

The  king  then  orders  another  pyramid,  of  the  same 
base,  of  which  altitude  is  to  be  a  mean  proportional  be- 
tween apothem  and  half-base — and  apothem  and  half- 
base  taken  as  one  line  are  to  be  in  mean  and  extreme 
ratio. 

The  architect's  plan  of  this  pyramid  will  be  the 
simple  figure  illustrated  by  me  (Fig.  70),  and  the 
dimensions  about — 


R.    B.   CUBITS. 


f  Base 420 

Base  angle  51°  49'  37-^?-    (  Altitude..  .267-1239849  &c. 

^  Apothem..  3397875 153  &c. 

But  the  builder  practically  carries   out  both  plans  when 
he  builds  to  my  templates  of  34  to  21  with — 


R.    B.  CUBITS. 


Base 420 

Base  angle  51°  51'  20"      <  Altitude  .  ..267*394839  &c. 

(^ Apothem.  .340 

and  neither  king  nor  architect  could  detect  error  in  the 
work. 

The  reader  will  remember  that  I  have  previously 
advanced  that  the  level  of  Cephren's  base  was  the  plan 
level  of  the  G'izeh  pyramids,  and  that  at  this  level  the 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  103 

base  of  Cheops  measures  420  R.B.  cubits — same  as  the 
base  of  Cephren. 

This  hypothesis  is  supported  by  the  revelations  of 
the  pentangle,  in  which  the  ratio  of  34  to  21  ==  apothem 
340  to  half-base  210  R.B.  cubits,  is  so  nearly  ap- 
proached. 

Showing  how  proportional  lines  were  the  order  of 
the  pyramids  of  Gizeh,  we  will   summarise  the   propor- 
tions of   the  three   main  pyramids  as   shewn  by  my  di- 
mensions and  ratios,  very  nearly,  viz.  : — 
Myicerinus.     Base  :  Apothem  :   :  Altitude  :  Half -Base. 

as  shown  by  the  ratios,  (Fig.  13),  40  :  32  ::   25:  20. 
Cephren.    Diagonal  of  Base  :  Edge  :   :  Edge  :  Altitude. 

as  shown  by  ratios,  (Fig.  12),  862  :  588   :  :  588  :  400. 
Cheops.     (Apothem  +  Half-Base}  :  Apoth.  :  :   Apoth.  : 
Half-Base. 

as  shown  by  the  ratios,  (Fig.  9),  55  :  34  :  :  34  121. 

and — Apothem  :  Altitude  :  :  Altitude  :  Half-B. 

Similar  close  relations  to  other  stars  may  be  found  in 
other  pyramids.  Thus  : — Suppose  NHO  of  figure  69  to 
be  the  NHO  of  a  heptangle  instead  of  a  pentangle,  then 
does  NH  represent  apothem,  and  NO  represent  base  of 
the  pyramid  Mycerinus,  while  the  co-sine  of  the  angle 
NHM  (being  MH  minus  versed  sine)  will  be  equal  to 
the  altitude  of  the  pyramid.  The  angle  NHM  in  the 
heptangle  is,  38°  34'  17*142",  and  according  to  my  plan 
of  the  pyramid  Mycerinus,  the  corresponding  angle  is 
38°  40'  56".  (See  Fig.  19.)  This  angular  difference  of 
o°  6'  39"  would  only  make  a  difference  in  the  apothem 


104  SOLUTION   OF  THE   PYRAMID   PROBLEM. 

of  the  pyramid  of  eight  inches,  and  of  ten  inches  in  its 
altitude  (apothem  being  283  //.  i  inch,  and  altitude 
22 1//.). 

§  17.     THE    MANNER    IN    WHICH    THE     SLOPE     RATIOS 
OF   THE    PYRAMIDS   WERE   ARRIVED    AT. 

The  manner  in  which  I  arrived  at  the  Slope  Ratios 
of  the  Pyramids,  viz.,  32  to  20,  33  to  20,  and  34  to  21, 
for  Myccrinus,  Ccphrcn,  and  Cheops,  respectively  (see 
Figures  8,  7  and  6),  was  as  follows  :— 

First,  believing  in  the  connection  between  the  rela- 
tive positions  of  the  Pyramids  on  plan  (see  Fig.  3,  4  or 
5),  and  their  slopes,  I  viewed  their  positions  thus  : — 

Mycerinus,  situate  at  the  angle  of  the  3,  4,  5  tri- 
angle ADC,  is  likely  to  be  connected  with  that  "pri- 
mary "  in  his  slopes. 

Cephren,  situate  at  the  angle  of  the  20,  21,  29  tri- 
angle FAB,  and  strung,  as  it  were,  on  the  hypotenuse 
of  the  3,  4,  5  triangle  DAC,  is  likely  to  be  connected 
with  both  primaries  in  his  slopes. 

Cheops,  situate  at  the  point  A,  common  to  both 
main  triangles,  governing  the  position  of  the  other 
pyramids,  is  likely  to  be  a  sort  of  mean  between  these 
two  pyramids  in  his  slope  ratios. 

Reasoning  thus,  with  the  addition  of  the  knowledge 
I  possessed  of  the  angular  estimates  of  these  slopes 
made  by  those  who  had  visited  the  ground,  and  a  use- 
ful start  for  my  ratios  gained  by  the  reduction  of  base 
measures  already  known  into  R.B.  cubits,  giving  420  as 


SOLUTION   OF   THE   PYRAMID   PROBLEM. 


105 


a  general  base  for  Cheops  and  Cephren  at  one  level, 
and  taking  210  cubits  as  the  base  of  Mycerinus  (half 
the  base  of  Cephren,  as  generally  admitted),  I  had 
something  solid  and  substantial  to  go  upon.  I  com- 
menced with  Mycerinus.  (See  Fig.  71.) 


(Mycerinus)    Fig.  71. 


LHNM  represents  the  base  of  the  pyramid.  On 
the  half-base  AC  I  described  a  3,  4,  5  triangle  ABC.  I 
then  projected  the  line  CF  ==  BC  to  be  the  altitude  of 
the  pyramid.  Thus  I  erected  the  triangle  BFC,  ratio  of 
BC  to  CF  being  i  to  i.  From  this  datum  I  arrived  at 
the  triangles  BEA,  ADC,  and  GKH.  GK,  EA,  and 
AD,  each  represent  apothem  of  pyramid  ;  CF,  and  CD, 
altitude  ;  and  H  K,  edge. 

The  length  of  the  line  AD  being  VAC2  +  CD2,  the 
length  of  the  line  HK  being  VHG2  +  GK2,  and  line  CH 
(half  diagonal  of  base)  being  VCG2  +  GH2.  These 
measures  reduced  to  R.B.  cubits,  calling  the  line  AC  = 
ratio  4  =:  105  cubits,  half-base  of  pyramid,  give  the  fol- 
lowing results  : — 


106  SOLUTION   OF   THE   PYRAMID    PROBLEM. 


R.    B.    CUBITS.      BRITISH  FEET. 


Half-base LA  ~  -  105*000      :  1 76*925 

Apothem EA  =  =  168*082       -  283*218 

Edge HK  =  :  198-183      =  333*937 

Altitude CD  =;   131*250     =  221*156 

Half  diag.  of  base..CH  -  :   148*4924  ==  250*209 
and  thus  I  acquired  the  ratios  :— 

Half-base  :  Altitude  :  :  Apothem  :  Base. 

=  20  :  25  :  :  32  :  40  nearly. 
To  place  the  lines  of  the  diagram  in  their  actual 
solid  position — Let  AB,  BC,  CA  and  HG  be  hinges 
attaching  the  planes  AEB,  BFC,  CD  A  and  HKG  to 
the  base  LHNM.  Lift  the  plane  BCF  on  its  hinge  till 
the  point  F  is  vertical  over  the  centre  C.  Lift  plane 
CDA  on  its  hinge,  till  point  D  is  vertical  over  the 
centre  C  ;  then  will  line  CD  touch  CF,  and  become 
one  line.  Now  lift  the  plane  AEB  on  its  hinge,  until 
point  E  is  vertical  over  the  centre  C,  and  plane  HKG 
on  its  hinge  till  point  K  is  vertical  over  the  centre  C  ; 
then  will  points  E,  F,  D  and  K,  all  meet  at  one  point 
above  the  centre  C,  and  all  the  lines  will  be  in  their 
proper  places. 

The  angle  at  the  base  of  Mycerinus,  if  built  to  a 
ratio  of  4  to  5  (half-base  to  altitude),  and  not  to  the 
more  practical  but  nearly  perfect  ratio  of  32  to  20  (apo- 
them  to  half-base)  would  be  the  complement  of  angle 
ADC,  thus- 

-i  =  '8  = :  Tan.  <  ADC    =  38°  39'  35^' 
5  477 

=  5i°  20'  24^- 

J  ^477 


SOLUTION   OF   THE    PYRAMID    PROBLEM. 


107 


but  as  it  is  probable  that  the  pyramid  w%s  buih  to  the 
ratio  of  32  to  20,  I  have  shown  its  base  angle  in  Figure 
19,  as  51°  19'  4". 

Figure  72  shows  how  the   slopes  of    Cephrcn   were 
arrived  at. 

(  Cephren  )    Fig.  72 , 


NT 


LHNM  represents  the  base  of  the  pyramid.  On 
the  half-base  AC,  I  described  a  3,  4,  5  triangle  ABC. 
I  then  projected  the  line  CF  (ratio  21  to  BC  20),  thus 
erecting  the  20,  21,  29  triangle  BCF.  From  this  datum, 
I  arrived  at  the  triangles  -BE A,  ADC,  and  GKH  ; 
GK,  EA  and  AD  each  representing  apothem  ;  CF  and 
CD,  altitude  ;  and  HK,  edge.  The  lengths  of  the  lines 
AD,  HK  and  CH  being  got  at  as  in  the  pyramid  My- 
cerinus.  These  measures  reduced  to  cubits,  calling 
AC  =  ratio  16  =  210  cubits  (half-base  of  pyramid)  give 
the  following  result. 


R.    B.    CUBITS. 


BRITISH    FEET. 


Half-base, 2  io'oo  353 ''8 5  —  LA 

Apothem 346*50  583*85  =  EA 

Edge .' 405*16  682-69  =  HK 

Altitude 275-625  464*43 '=  CD 

Half-diag.of  base  296-985  5oo*42  =  CH 


io8 


SOLUTION  OF  THE  PYRAMID  PROBLEM. 


thus  I  get  the  ratios  of — Apothem  :  Half-Base  :  :  33  : 
20,  &c.  The  planes  in  the  diagram  are  placed  in  their 
correct  positions,  as  directed  for  Figure  71. 

The  angle  at  the  base  of  Cephren,  if  built  to  the 
ratio  of  1 6  to  21  (half-base  to  altitude),  and  not  to  the 
practical  ratio  of  33  to  20  (apothem  to  half-base),  would 
be  the  complement  of  <  ADC,  thus— 

1(3  -    761904  =     Tan.  <  ADC -37°  18'   14!!" 

:  DAC  -  52°  4i/45/'^- 

but  as  it  is  probable  that  the  pyramid  was  built  to  the 
ratio  of  33  to  20,  I  have  marked  the  base  angle  in  Fig. 
17,  as  52°  41'  41". 

I  took  Cheops  out,  first  as  a  K  pyramid,  and  made 
his  lines  to  a  base  of  420  cubits,  as  follows— 

Half-base 210 

Altitude 267.380304 

Apothem 339-988573  (See  Fig.  73.) 

(Cheops)     Fig.  T3. 


But  to  produce   the  building  ratio  of  34  to  21,  as  per 
diagram  Figure  6  or  9,  I  had  to  alter  it  to — 


SOLUTION   OF  THE   PYRAMID   PROBLEM.  109 

Half-base 210 

Altitude 267-394839 

Apothem 340 

Thus  the  theoretical  angle  of  Cheops  is  51°  51'  H'3", 
and  the  probable  angle  at  which  it  was  built,  is  51°  51' 
20",  as  per  figure  15. 

Cheops  is  therefore  the  mean  or  centre  of  a  system— 
the  slopes  of  Mycerinus  being  a  little  flatter,  and  those 
of  Cephren  a  little  steeper,  Cheops  coming  fairly  be- 
tween the  two,  within  about  10  minutes  ;  and  thus  the 
connection  between  the  ground  plan  of  the  group  and 
the  slopes  of  the  three  pyramids  is  exactly  as  one  might 
expect  after  examination  of  Figure  3,  4  or  5. 


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